0 added
0 removed
Original
2026-01-01
Modified
2026-02-28
1
<p>We can derive the derivative of 5tanx using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation.</p>
1
<p>We can derive the derivative of 5tanx using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation.</p>
2
<p>There are several methods we use to prove this, such as:</p>
2
<p>There are several methods we use to prove this, such as:</p>
3
<ul><li>By First Principle </li>
3
<ul><li>By First Principle </li>
4
<li>Using Chain Rule</li>
4
<li>Using Chain Rule</li>
5
<li>Using Product Rule</li>
5
<li>Using Product Rule</li>
6
</ul><p>We will now demonstrate that the differentiation of 5tanx results in 5sec²x using the above-mentioned methods:</p>
6
</ul><p>We will now demonstrate that the differentiation of 5tanx results in 5sec²x using the above-mentioned methods:</p>
7
<h2>By First Principle</h2>
7
<h2>By First Principle</h2>
8
<p>The derivative of 5tanx can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of 5tanx using the first principle, we will consider f(x) = 5tanx. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 5tanx, we write f(x + h) = 5tan(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [5tan(x + h) - 5tanx] / h = 5 limₕ→₀ [tan(x + h) - tanx] / h = 5 limₕ→₀ [[sin(x + h) / cos(x + h)] - [sinx / cosx]] / h = 5 limₕ→₀ [[sin(x + h)cosx - cos(x + h)sinx] / [cosx · cos(x + h)]] / h We now use the formula sinAcosB - cosAsinB = sin(A-B). f'(x) = 5 limₕ→₀ [sinh] / [h cosx · cos(x + h)] = 5 limₕ→₀ [sinh] / [h cosx · cos(x + h)] = 5 limₕ→₀ (sinh)/h · limₕ→₀ 1 / [cosx · cos(x + h)] Using limit formulas, limₕ→₀ (sinh)/h = 1. f'(x) = 5 [1 / (cosx · cos(x + 0))] = 5/cos²x As the reciprocal of cosine is secant, we have, f'(x) = 5sec²x. Hence, proved.</p>
8
<p>The derivative of 5tanx can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of 5tanx using the first principle, we will consider f(x) = 5tanx. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 5tanx, we write f(x + h) = 5tan(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [5tan(x + h) - 5tanx] / h = 5 limₕ→₀ [tan(x + h) - tanx] / h = 5 limₕ→₀ [[sin(x + h) / cos(x + h)] - [sinx / cosx]] / h = 5 limₕ→₀ [[sin(x + h)cosx - cos(x + h)sinx] / [cosx · cos(x + h)]] / h We now use the formula sinAcosB - cosAsinB = sin(A-B). f'(x) = 5 limₕ→₀ [sinh] / [h cosx · cos(x + h)] = 5 limₕ→₀ [sinh] / [h cosx · cos(x + h)] = 5 limₕ→₀ (sinh)/h · limₕ→₀ 1 / [cosx · cos(x + h)] Using limit formulas, limₕ→₀ (sinh)/h = 1. f'(x) = 5 [1 / (cosx · cos(x + 0))] = 5/cos²x As the reciprocal of cosine is secant, we have, f'(x) = 5sec²x. Hence, proved.</p>
9
<h2>Using Chain Rule</h2>
9
<h2>Using Chain Rule</h2>
10
<p>To prove the differentiation of 5tanx using the chain rule, We use the formula: Tanx = sinx/cosx Consider f(x) = sinx and g(x) = cosx So we get, tanx = f(x)/g(x) By quotient rule: d/dx [f(x) / g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]² … (1) Let’s substitute f(x) = sinx and g(x) = cosx in equation (1), d/dx (tanx) = [(cosx)(cosx) - (sinx)(-sinx)] / (cosx)² (cos²x + sin²x) / cos²x …(2) Here, we use the formula: (cos²x) + (sin²x) = 1 (Pythagorean identity) Substituting this into (2), d/dx (tanx) = 1 / (cosx)² Since secx = 1/cosx, we write: d/dx(tanx) = sec²x Thus, d/dx (5tanx) = 5sec²x</p>
10
<p>To prove the differentiation of 5tanx using the chain rule, We use the formula: Tanx = sinx/cosx Consider f(x) = sinx and g(x) = cosx So we get, tanx = f(x)/g(x) By quotient rule: d/dx [f(x) / g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]² … (1) Let’s substitute f(x) = sinx and g(x) = cosx in equation (1), d/dx (tanx) = [(cosx)(cosx) - (sinx)(-sinx)] / (cosx)² (cos²x + sin²x) / cos²x …(2) Here, we use the formula: (cos²x) + (sin²x) = 1 (Pythagorean identity) Substituting this into (2), d/dx (tanx) = 1 / (cosx)² Since secx = 1/cosx, we write: d/dx(tanx) = sec²x Thus, d/dx (5tanx) = 5sec²x</p>
11
<h2>Using Product Rule</h2>
11
<h2>Using Product Rule</h2>
12
<p>We will now prove the derivative of 5tanx using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, 5tanx = 5(sinx/cosx) 5tanx = 5(sinx)(cosx)⁻¹ Given that, u = sinx and v = (cosx)⁻¹ Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (sinx) = cosx. (substitute u = sinx) Here we use the chain rule: v = (cosx)⁻¹ = (cosx)⁻¹ (substitute v = (cosx)⁻¹) v' = -1(cos)⁻² d/dx (cosx) v' = sinx / (cosx)² Again, use the product rule formula: d/dx (5tanx) = 5[u'.v + u.v'] Let’s substitute u = sinx, u' = cosx, v = (cosx)⁻¹, and v' = sinx / (cosx)² When we simplify each<a>term</a>: We get, d/dx (5tanx) = 5(1 + sin²x / (cosx)²) sin²x / (cosx)² = tan²x (we use the identity sin²x + cos²x = 1) Thus: d/dx (5tanx) = 5(1 + tan²x) Since, 1 + tan²x = sec²x d/dx (5tanx) = 5sec²x.</p>
12
<p>We will now prove the derivative of 5tanx using the<a>product</a>rule. The step-by-step process is demonstrated below: Here, we use the formula, 5tanx = 5(sinx/cosx) 5tanx = 5(sinx)(cosx)⁻¹ Given that, u = sinx and v = (cosx)⁻¹ Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (sinx) = cosx. (substitute u = sinx) Here we use the chain rule: v = (cosx)⁻¹ = (cosx)⁻¹ (substitute v = (cosx)⁻¹) v' = -1(cos)⁻² d/dx (cosx) v' = sinx / (cosx)² Again, use the product rule formula: d/dx (5tanx) = 5[u'.v + u.v'] Let’s substitute u = sinx, u' = cosx, v = (cosx)⁻¹, and v' = sinx / (cosx)² When we simplify each<a>term</a>: We get, d/dx (5tanx) = 5(1 + sin²x / (cosx)²) sin²x / (cosx)² = tan²x (we use the identity sin²x + cos²x = 1) Thus: d/dx (5tanx) = 5(1 + tan²x) Since, 1 + tan²x = sec²x d/dx (5tanx) = 5sec²x.</p>
13
13