Find the derivative: frac{d}{dx}[cos((1 - x^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To find the derivative of $\cos((1 - x^2)^2)$,we use the chain rule.Let $y = \cos((1 - x^2)^2)$.Applying the chain rule: $\frac{dy}{dx} = -\sin((1

Vedclass · Accepted Answer

$4x(1 - x^2)\sin((1 - x^2)^2)$

Vedclass · Answer

(C) To find the derivative of $\cos((1 - x^2)^2)$,we use the chain rule.Let $y = \cos((1 - x^2)^2)$.Applying the chain rule: $\frac{dy}{dx} = -\sin((1 - x^2)^2) \cdot \frac{d}{dx}((1 - x^2)^2)$.Now,differentiate $(1 - x^2)^2$ using the power rule and chain rule:$\frac{d}{dx}((1 - x^2)^2) = 2(1 - x^2) \cdot \frac{d}{dx}(1 - x^2) = 2(1 - x^2) \cdot (-2x) = -4x(1 - x^2)$.Substituting this back into the expression:$\frac{dy}{dx} = -\sin((1 - x^2)^2) \cdot (-4x(1 - x^2)) = 4x(1 - x^2)\sin((1 - x^2)^2)$.Thus,the correct option is $C$.

Find the derivative: $\frac{d}{dx}[\cos((1 - x^2)^2)] = ?$

Similar Questions

Find the derivative: $\frac{d}{dx} \left( \frac{\log x}{\sin x} \right)$

Differentiate the following with respect to $x$: $e^{x^{3}}$

The derivative of $\sin \left( \log \left( \frac{x + 3}{x} \right) \right)$ with respect to $x$ is:

At $x = \sqrt{\frac{\pi}{2}}$,find the value of $\frac{d}{dx} \cos(\sin(x^2))$.

Let $f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}}$ for $1 < x < 26$ be a real-valued function. Then $f'(x)$ for $1 < x < 26$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module