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

If $\frac{d}{dx} \left( \frac{x^4 + x^2 + 1}{x^2 + x + 1} \right) = ax + b$,then $a - b =$

Let $f(x)$ be a polynomial function such that $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^{5}+64$. Then,the value of $\lim _{x \rightarrow 1} \frac{f(x)}{x-1}$ is:

The first derivative of the function $(\sin 2x \cos 2x \cos 3x + \log_2 2^{x+3})$ with respect to $x$ at $x = \pi$ is

Let $f:(-1,1) \rightarrow \mathbb{R}$ be a differentiable function with $f(0)=-1$ and $f^{\prime}(0)=1$. If $g(x)=(f(2f(x)+2))^2$, then $g^{\prime}(0)=$

If $f(x) = \sqrt{x + 2 \sqrt{2x - 4}} + \sqrt{x - 2 \sqrt{2x - 4}}$,then the value of $10 \times f^{\prime}(102) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module