Question

If f(x)=3sinx+4tanx. Find f'(π).​

Answer 1

Topic - Differentiation

Explanation:

We need to find the derivative of f(x) at x = π for the function $\rm f(x) = 3\sin(x) + 4\tan(x)$.

We will use the following formulas to find the derivative.

$\boxed{\begin{array}{lc}\rm \dfrac{d}{dx}\{f(x) \pm g(x) \} = \dfrac{d}{dx}\{f(x)\}\pm\dfrac{d}{dx}\{g(x)\}\\\\\rm\dfrac{d}{dx}\{kf(x)\} = k\dfrac{d}{dx}\{f(x)\}\\\\\rm\dfrac{d}{dx}\{\tan(x)\} = \sec^2(x)\\\\\rm\dfrac{d}{dx}\{\sin(x)\}= \cos(x)\end{array}}$

By using  these formulas, derivative of the given function is given by,

$\sf\implies f'(x) = \dfrac{d}{dx}\{3\sin(x) + 4\tan(x)\}$

$\sf\implies f'(x) = \dfrac{d}{dx}\{3\sin(x) \}+\dfrac{d}{dx}\{ 4\tan(x)\}$

$\sf\implies f'(x) = 3\dfrac{d}{dx}\{\sin(x) \}+4\dfrac{d}{dx}\{\tan(x)\}$

$\sf\implies f'(x) = 3\cos(x)+4\sec^2(x)$

Now, the derivative of f(x) at x = π is given by,

$\sf\implies f'(\pi) = 3\cos(\pi)+4\sec^2(\pi)$

$\sf\implies f'(\pi) = 3(-1)+4(-1)^2$

$\sf\implies f'(\pi) = -3+4$

$\sf\implies f'(\pi) = 1$

Hence the value of f'(π) is 1.