Derivative of 1/cos(x)
2026-02-28 12:06 Diff
https://brightchamps.com/en-us/math/calculus/derivative-of-1-by-cosx

We can derive the derivative of 1/cos(x) using proofs.

To show this, we will use the trigonometric identities along with the rules of differentiation.

There are several methods we use to prove this, such as:

By First Principle Using Chain Rule Using Product Rule We will now demonstrate that the differentiation of 1/cos(x) results in sec(x)tan(x) using the above-mentioned methods: By First Principle The derivative of 1/cos(x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.

To find the derivative of 1/cos(x) using the first principle, we will consider f(x) = 1/cos(x).

Its derivative can be expressed as the following limit.

f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)

Given that f(x) = 1/cos(x), we write f(x + h) = 1/cos(x + h).

Substituting these into equation (1), f'(x) = limₕ→₀ [1/cos(x + h) - 1/cos(x)] / h = limₕ→₀ [(cos(x) - cos(x + h)) / (cos(x)cos(x + h))] / h = limₕ→₀ [(-sin(x + h/2)sin(h/2)) / (cos(x)cos(x + h))]/ h

Using the identity for sin A - sin B, f'(x) = limₕ→₀ [-sin(x + h/2)sin(h/2)] / [h cos(x)cos(x + h)] = limₕ→₀ [-sin(x + h/2)(h/2)/(h/2)cos(x)cos(x + h)]

Using limit formula, limₕ→₀ (sin(h/2))/(h/2) = 1.

f'(x) = -sin(x) / cos²(x) As the reciprocal of cosine is secant, and using tan(x) = sin(x)/cos(x), we have, f'(x) = sec(x)tan(x). Hence, proved.

Using Chain Rule To prove the differentiation of 1/cos(x) using the chain rule, Consider y = 1/cos(x).

Let u = cos(x), then y = 1/u.

Using the chain rule: dy/dx = dy/du * du/dx.

We have dy/du = -1/u² and du/dx = -sin(x).

Therefore, dy/dx = -1/cos²(x) * -sin(x) = sin(x)/cos²(x).

Using the identity tan(x) = sin(x)/cos(x), we have, dy/dx = sec(x)tan(x). Using Product Rule We will now prove the derivative of 1/cos(x) using the product rule.

Let y = sec(x) = 1/cos(x) = (cos(x))⁻¹.

Using the product rule formula: d/dx [u.v] = u'.v + u.v'. Let u = 1 and v = (cos(x))⁻¹.

Then v' = -sin(x)/(cos(x))². Using the product rule, dy/dx = 0 + 1 * (-sin(x)/(cos(x))²) = -sin(x)/cos²(x) = sec(x)tan(x).