Find the derivative of the following function | vedclass

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

Question

(A) Let $f(x) = 3 \cot x + 5 \operatorname{cosec} x$. Using the derivative rules:$\frac{d}{dx} [3 \cot x + 5 \operatorname{cosec} x] = 3 \frac{d}{dx}

Vedclass · Accepted Answer

$-3 \operatorname{cosec}^{2} x - 5 \operatorname{cosec} x \cot x$

Vedclass · Answer

(A) Let $f(x) = 3 \cot x + 5 \operatorname{cosec} x$. Using the derivative rules:$\frac{d}{dx} [3 \cot x + 5 \operatorname{cosec} x] = 3 \frac{d}{dx} (\cot x) + 5 \frac{d}{dx} (\operatorname{cosec} x)$We know that $\frac{d}{dx} (\cot x) = -\operatorname{cosec}^{2} x$ and $\frac{d}{dx} (\operatorname{cosec} x) = -\operatorname{cosec} x \cot x$.Substituting these values:$f'(x) = 3(-\operatorname{cosec}^{2} x) + 5(-\operatorname{cosec} x \cot x)$$f'(x) = -3 \operatorname{cosec}^{2} x - 5 \operatorname{cosec} x \cot x$

Find the derivative of the following function: $3 \cot x + 5 \operatorname{cosec} x$

Similar Questions

If $f$ and $g$ are differentiable functions satisfying $g^{\prime}(a)=2$,$g(a)=b$,and $f \circ g = I$,where $I$ is an identity function,then $f^{\prime}(b)$ is equal to

Find the derivative of $f$ given by $f(x) = \sin^{-1} x$ assuming it exists.

Let $f(x)$ be a polynomial of degree $3$ such that $f(3)=1$,$f'(3) = -1$,$f''(3) = 0$ and $f'''(3)=12$. Then the value of $f'(1)$ is

Find the derivative of the following function (it is to be understood that $p, q, r$ are fixed non-zero constants): $\frac{p x^{2}+q x+r}{x}$

If $y = x^2 + \cos(2x) + e^{ax}$,then find $\frac{dy}{dx}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module