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2026-01-01
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2026-02-28
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<p>157 Learners</p>
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<p>187 Learners</p>
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<p>Last updated on<strong>September 26, 2025</strong></p>
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<p>Last updated on<strong>September 26, 2025</strong></p>
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<p>An implicit function describes the relationship between independent variables and a dependent variable. In a single equation, like y - 3x2 + 2x + 5 = 0, an explicit function shows the output clearly, but an implicit function doesn’t. Like y = 3x + 2. Implicit differentiation uses the chain rule or product rule to find the derivative without solving for y.</p>
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<p>An implicit function describes the relationship between independent variables and a dependent variable. In a single equation, like y - 3x2 + 2x + 5 = 0, an explicit function shows the output clearly, but an implicit function doesn’t. Like y = 3x + 2. Implicit differentiation uses the chain rule or product rule to find the derivative without solving for y.</p>
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<h2>What Is An Implicit Function?</h2>
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<h2>What Is An Implicit Function?</h2>
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<p>What Is Algebra? 🧮 | Simple Explanation with 🎯 Cool Examples for Kids | ✨BrightCHAMPS Math</p>
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<p>What Is Algebra? 🧮 | Simple Explanation with 🎯 Cool Examples for Kids | ✨BrightCHAMPS Math</p>
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<p>▶</p>
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<p>▶</p>
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<p>An implicit<a>function</a>includes the dependent<a>variable</a>within an<a>equation</a>, like x2+y2=1, without solving for it. Instead<a>of</a>giving y explicitly as y = f(x). Implicit differentiation lets you find dy/dx by differentiating both sides using the chain rule without solving for y first. Implicit differentiation is useful when it’s hard or impossible to solve for y directly. </p>
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<p>An implicit<a>function</a>includes the dependent<a>variable</a>within an<a>equation</a>, like x2+y2=1, without solving for it. Instead<a>of</a>giving y explicitly as y = f(x). Implicit differentiation lets you find dy/dx by differentiating both sides using the chain rule without solving for y first. Implicit differentiation is useful when it’s hard or impossible to solve for y directly. </p>
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<h2>Properties of Implicit Function</h2>
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<h2>Properties of Implicit Function</h2>
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<p>Cannot be written as y = f(x)</p>
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<p>Cannot be written as y = f(x)</p>
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<p>The dependent variable y isn't isolated; both x and y appear together in the same equation</p>
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<p>The dependent variable y isn't isolated; both x and y appear together in the same equation</p>
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<p>Always in the form f(x, y) = 0</p>
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<p>Always in the form f(x, y) = 0</p>
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<p>The equation combines x and y like x2+y2-1 = 0 for a circle</p>
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<p>The equation combines x and y like x2+y2-1 = 0 for a circle</p>
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<p>Frequently non-linear and multivariable</p>
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<p>Frequently non-linear and multivariable</p>
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<p>Many implicit functions are<a>polynomials</a>or complex relations, including many variables</p>
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<p>Many implicit functions are<a>polynomials</a>or complex relations, including many variables</p>
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<p>Combines dependent and independent<a>terms</a>.</p>
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<p>Combines dependent and independent<a>terms</a>.</p>
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<p>These cannot be easily solved for y because it stays within the equation with x.</p>
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<p>These cannot be easily solved for y because it stays within the equation with x.</p>
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<p>It may not pass the<a>vertical line test</a>.</p>
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<p>It may not pass the<a>vertical line test</a>.</p>
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<p>An implicit relation does not always pass the vertical line test because: A vertical line might cross its graph more than once. That means y is not a single-valued function of x. </p>
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<p>An implicit relation does not always pass the vertical line test because: A vertical line might cross its graph more than once. That means y is not a single-valued function of x. </p>
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<h2>Derivative of Implicit Function</h2>
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<h2>Derivative of Implicit Function</h2>
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<p>While solving an equation involving both x and y together f(x, y) = 0, you can't isolate y very easily, you directly differentiate both sides without solving a variable first. With respect to x treating y as a function of x and use the chain rule whenever you differentiate a term including y.</p>
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<p>While solving an equation involving both x and y together f(x, y) = 0, you can't isolate y very easily, you directly differentiate both sides without solving a variable first. With respect to x treating y as a function of x and use the chain rule whenever you differentiate a term including y.</p>
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<p>Differentiate each term in the equation with respect to x, adding dydx while differentiating anything with y Collect all dydx, terms on one side Solve for dydx, the result will usually involve both x and y</p>
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<p>Differentiate each term in the equation with respect to x, adding dydx while differentiating anything with y Collect all dydx, terms on one side Solve for dydx, the result will usually involve both x and y</p>
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<p>x2+xy+y=0</p>
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<p>x2+xy+y=0</p>
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<p>Differentiate both sides w.r.t.x:</p>
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<p>Differentiate both sides w.r.t.x:</p>
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<p>ddx(x2)+ddx(xy)+ddx(y)=0</p>
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<p>ddx(x2)+ddx(xy)+ddx(y)=0</p>
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<p>Becomes</p>
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<p>Becomes</p>
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<p>2x+(xdydx+y1)+dydx=0</p>
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<p>2x+(xdydx+y1)+dydx=0</p>
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<p>Combine dydx terms and isolate</p>
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<p>Combine dydx terms and isolate</p>
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<p>(x+1)dydx+(2x+y)=0dydx=-2x+yx+1</p>
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<p>(x+1)dydx+(2x+y)=0dydx=-2x+yx+1</p>
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<p>You get dydx directly without needing to solve for y first. </p>
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<p>You get dydx directly without needing to solve for y first. </p>
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<h2>Implicit Function Theorem</h2>
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<h2>Implicit Function Theorem</h2>
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<p>The implicit function theorem helps to turn complicated relationships into simpler functions of real variables. It shows that, even if a relation doesn’t define a function globally, it can behave like a function locally. For example, given an equation like F(x, y)=0, the theorem ensures that we can solve for y as a smooth function of x. Certain partial derivatives are non-zero, indicating how the function changes as x changes. </p>
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<p>The implicit function theorem helps to turn complicated relationships into simpler functions of real variables. It shows that, even if a relation doesn’t define a function globally, it can behave like a function locally. For example, given an equation like F(x, y)=0, the theorem ensures that we can solve for y as a smooth function of x. Certain partial derivatives are non-zero, indicating how the function changes as x changes. </p>
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<h2>Real-Life Applications of the Implicit Function</h2>
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<h2>Real-Life Applications of the Implicit Function</h2>
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<p>Real life application is important in many fields such as economics, physics, chemistry, The uses are explained below;</p>
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<p>Real life application is important in many fields such as economics, physics, chemistry, The uses are explained below;</p>
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<ul><li>Economics-marginal<a>rate</a>of substitution: Economists use implicit differentiation to analyze trade-offs between goods to find how much more of one good is needed to remain equally satisfied when consuming less of another, for example for utility equation U(x, y)=k, implicit differentiate gives dy/dx = -(Ux/Uy)</li>
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<ul><li>Economics-marginal<a>rate</a>of substitution: Economists use implicit differentiation to analyze trade-offs between goods to find how much more of one good is needed to remain equally satisfied when consuming less of another, for example for utility equation U(x, y)=k, implicit differentiate gives dy/dx = -(Ux/Uy)</li>
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<li>Ladder sliding problem: This is used by teachers and students to solve motion problems in physics. As a ladder slides down a wall, its top moves downward while the<a>base</a>slides outward. To relate the rates of change of these positions, we use implicit differentiation. For example, if the ladder has a fixed length h, and its position forms a right triangle, the relationship is: x2+y2=h2 </li>
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<li>Ladder sliding problem: This is used by teachers and students to solve motion problems in physics. As a ladder slides down a wall, its top moves downward while the<a>base</a>slides outward. To relate the rates of change of these positions, we use implicit differentiation. For example, if the ladder has a fixed length h, and its position forms a right triangle, the relationship is: x2+y2=h2 </li>
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<li>By differentiating both sides with respect to time t, we can find dydt (how fast the top is falling) in terms of dxdt (how fast the base is sliding).</li>
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<li>By differentiating both sides with respect to time t, we can find dydt (how fast the top is falling) in terms of dxdt (how fast the base is sliding).</li>
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<li>Physics-motion with constraints: Physicists analyzing moving or constrained systems often deal with motion equations in implicit form. Implicit differentiation is used to find quantities ;like velocity and acceleration when dependent on<a>multiple</a>variables. For example, pendulum constraints x2+y2=l2; differentiate implicitly to find angular acceleration</li>
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<li>Physics-motion with constraints: Physicists analyzing moving or constrained systems often deal with motion equations in implicit form. Implicit differentiation is used to find quantities ;like velocity and acceleration when dependent on<a>multiple</a>variables. For example, pendulum constraints x2+y2=l2; differentiate implicitly to find angular acceleration</li>
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<li>Fluid dynamics/ engineering: This is used by engineers modeling pressure and flows, where equations relating to pressure, velocity, and<a>geometry</a>are often implicit. Implicit differentiation helps to find the gradients, such as in the Navier-stokes equation, to predict the fluid behavior.</li>
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<li>Fluid dynamics/ engineering: This is used by engineers modeling pressure and flows, where equations relating to pressure, velocity, and<a>geometry</a>are often implicit. Implicit differentiation helps to find the gradients, such as in the Navier-stokes equation, to predict the fluid behavior.</li>
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<li>Thermodynamics/ chemistry: Implicit differentiation is used in thermodynamics. This helps in finding how the pressure changes with volume or temperature in gas laws like the Van der Waals equation, where variables are interdependent. </li>
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<li>Thermodynamics/ chemistry: Implicit differentiation is used in thermodynamics. This helps in finding how the pressure changes with volume or temperature in gas laws like the Van der Waals equation, where variables are interdependent. </li>
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</ul><h2>Common Mistakes With Implicit Function and How to Avoid Them</h2>
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</ul><h2>Common Mistakes With Implicit Function and How to Avoid Them</h2>
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<p>While solving implicit function, students often forget that y depends on x in implicit differentiation. </p>
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<p>While solving implicit function, students often forget that y depends on x in implicit differentiation. </p>
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<h2>Download Worksheets</h2>
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<h3>Problem 1</h3>
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<h3>Problem 1</h3>
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<p>Differentiate the equation x2+y2=2.</p>
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<p>Differentiate the equation x2+y2=2.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p> dydx=-xy </p>
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<p> dydx=-xy </p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>Here, we are given a relation involving both x and y. Since y is implicit of a function x, we apply the chain rule while differentiating y2. The derivative of the x2 is 2x, and the derivative of y2 is 2ydydx. After the differentiation, we only take dydx to get the final answer. ddx(x2+y2)=ddx(2) 2x+2ydydx=0 dydx=-xy </p>
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<p>Here, we are given a relation involving both x and y. Since y is implicit of a function x, we apply the chain rule while differentiating y2. The derivative of the x2 is 2x, and the derivative of y2 is 2ydydx. After the differentiation, we only take dydx to get the final answer. ddx(x2+y2)=ddx(2) 2x+2ydydx=0 dydx=-xy </p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 2</h3>
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<h3>Problem 2</h3>
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<p>Differentiate the equation x2+y3=4</p>
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<p>Differentiate the equation x2+y3=4</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>dydx=-2x3y2 </p>
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<p>dydx=-2x3y2 </p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>Differentiate both the sides of the equation with respect to x. Apply the power rule to x2 and the chain rule to y3(since y is a function of x). After differentiating, we rearrange the equation to isolate dydx. ddx(x2+y3)=ddx(4) 2x+3y2dydx=0 Then, dydx=-2x3y2 </p>
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<p>Differentiate both the sides of the equation with respect to x. Apply the power rule to x2 and the chain rule to y3(since y is a function of x). After differentiating, we rearrange the equation to isolate dydx. ddx(x2+y3)=ddx(4) 2x+3y2dydx=0 Then, dydx=-2x3y2 </p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 3</h3>
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<h3>Problem 3</h3>
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<p>Find the derivative dydx for the equation x . y3=1</p>
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<p>Find the derivative dydx for the equation x . y3=1</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>dydx=-y3x </p>
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<p>dydx=-y3x </p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>Here, we use the product rule to differentiate x . y3. The derivative of x is said to be 1 and that of y3 is 3y2dydx. Now, simplify to separate dydx. Finally, reduce the expression. ddx(xy3)=ddx(1) y3+x . 3y2dydx=0 3xy2dydx=-y3 dydx=-y3x</p>
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<p>Here, we use the product rule to differentiate x . y3. The derivative of x is said to be 1 and that of y3 is 3y2dydx. Now, simplify to separate dydx. Finally, reduce the expression. ddx(xy3)=ddx(1) y3+x . 3y2dydx=0 3xy2dydx=-y3 dydx=-y3x</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 4</h3>
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<h3>Problem 4</h3>
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<p>Differentiate the equation 2y3+4x2-y=x6</p>
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<p>Differentiate the equation 2y3+4x2-y=x6</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p> dydx=6x5-8x6y2-1 </p>
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<p> dydx=6x5-8x6y2-1 </p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p> Here, differentiate each term of the equation. Use the chain rule for y3 and y, and the power rule for the x2 and x6. Now, group the terms containing dydx, factor it out, and solve for it. Here, ddx(2y3+4x2-y)=ddx(x6) 6y2dydx+8x-dydx=6x5 (6y2-1)dydx=6x5-8x dydx=6x5-8x6y2-1 </p>
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<p> Here, differentiate each term of the equation. Use the chain rule for y3 and y, and the power rule for the x2 and x6. Now, group the terms containing dydx, factor it out, and solve for it. Here, ddx(2y3+4x2-y)=ddx(x6) 6y2dydx+8x-dydx=6x5 (6y2-1)dydx=6x5-8x dydx=6x5-8x6y2-1 </p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 5</h3>
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<h3>Problem 5</h3>
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<p>Differentiate x2+xy+y2=4</p>
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<p>Differentiate x2+xy+y2=4</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>dydx=-2x+yx+2y </p>
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<p>dydx=-2x+yx+2y </p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>Here, to differentiate xy we need the product rule, which is derivative of x is 1, and derivative of y is dydx. After applying the derivative to each term, combine the like terms and isolate dydx. Here, ddx(x2+xy+y2)=ddx(4) 2x+(xdydx+y)+2ydydx=0 (x+2y)dydx=-2x-y dydx=-2x-yx+2y </p>
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<p>Here, to differentiate xy we need the product rule, which is derivative of x is 1, and derivative of y is dydx. After applying the derivative to each term, combine the like terms and isolate dydx. Here, ddx(x2+xy+y2)=ddx(4) 2x+(xdydx+y)+2ydydx=0 (x+2y)dydx=-2x-y dydx=-2x-yx+2y </p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h2>FAQs of the Implicit Function</h2>
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<h2>FAQs of the Implicit Function</h2>
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<h3>1. What is the implicit function?</h3>
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<h3>1. What is the implicit function?</h3>
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<p>A function that cannot be easily represented in the form y=f(x) is called an implicit function. </p>
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<p>A function that cannot be easily represented in the form y=f(x) is called an implicit function. </p>
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<h3>2.What is an implicit type function?</h3>
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<h3>2.What is an implicit type function?</h3>
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<p>An implicit type function is a function where the dependent variable is not isolated. But, both the dependent and independent variables appear together in a single equation. </p>
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<p>An implicit type function is a function where the dependent variable is not isolated. But, both the dependent and independent variables appear together in a single equation. </p>
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<h3>3.What is the explicit function?</h3>
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<h3>3.What is the explicit function?</h3>
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<p> A function that is often written as one variable, or dependent variable, in terms of another variable, or independent variable.</p>
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<p> A function that is often written as one variable, or dependent variable, in terms of another variable, or independent variable.</p>
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<h3>4.What is an example of implicit type?</h3>
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<h3>4.What is an example of implicit type?</h3>
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<p>Multiplying an<a>integer</a>by a float, the integer will be prompted to a float for the duration of the evaluation of the<a>expression</a>. </p>
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<p>Multiplying an<a>integer</a>by a float, the integer will be prompted to a float for the duration of the evaluation of the<a>expression</a>. </p>
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<h3>5.What is called implicit differentiation?</h3>
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<h3>5.What is called implicit differentiation?</h3>
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<p>Implicit differentiation is the process of finding the derivatives when the dependent variable is not isolated in the equation. </p>
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<p>Implicit differentiation is the process of finding the derivatives when the dependent variable is not isolated in the equation. </p>
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