Differentiate the function with respect to x: | vedclass

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

Question

(B) Let $y = \cos (a \cos x + b \sin x)$.By using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} [\cos (a \cos x +

Vedclass · Accepted Answer

$(a \sin x - b \cos x) \sin (a \cos x + b \sin x)$

Vedclass · Answer

(B) Let $y = \cos (a \cos x + b \sin x)$.By using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} [\cos (a \cos x + b \sin x)]$Using the derivative of $\cos(u)$ which is $-\sin(u) \cdot \frac{du}{dx}$:$\frac{dy}{dx} = -\sin (a \cos x + b \sin x) \cdot \frac{d}{dx} (a \cos x + b \sin x)$Now,differentiate the inner function:$\frac{d}{dx} (a \cos x + b \sin x) = a(-\sin x) + b(\cos x) = b \cos x - a \sin x$Substituting this back into the expression:$\frac{dy}{dx} = -\sin (a \cos x + b \sin x) \cdot (b \cos x - a \sin x)$Rearranging the terms:$\frac{dy}{dx} = (a \sin x - b \cos x) \sin (a \cos x + b \sin x)$

Differentiate the function with respect to $x$: $\cos (a \cos x + b \sin x)$,where $a$ and $b$ are constants.

Similar Questions

Differentiate the following with respect to $x$: $\sin \left(\tan ^{-1} e^{-x}\right)$

Let $f: (-1, 1) \to R$ be a differentiable function with $f(0) = -1$ and $f'(0) = 1$. Let $g(x) = [f(2f(x) + 2)]^2$. Then $g'(0) = $

If $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$,then $\frac{dy}{dx}$ equals:

If $y = \frac{(a - x)\sqrt{a - x} - (b - x)\sqrt{x - b}}{\sqrt{a - x} + \sqrt{x - b}}$,then $\frac{dy}{dx}$ wherever it is defined is equal to:

If $y = \sin(2 \sin^{-1} x)$, then $\frac{dy}{dx} = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module