frac{d}{d x}left(frac{x+5}{(x+1)^2(x+2)}right | vedclass

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

Question

(B) Let $f(x) = \frac{x+5}{(x+1)^2(x+2)}$. Using partial fractions, we write: $\frac{x+5}{(x+1)^2(x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}

Vedclass · Accepted Answer

$\frac{3}{(x+1)^2}-\frac{3}{(x+2)^2}-\frac{8}{(x+1)^3}$

Vedclass · Answer

(B) Let $f(x) = \frac{x+5}{(x+1)^2(x+2)}$. Using partial fractions, we write: $\frac{x+5}{(x+1)^2(x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2}$ $x+5 = A(x+1)(x+2) + B(x+2) + C(x+1)^2$ For $x = -1$: $-1+5 = B(-1+2) \Rightarrow B = 4$. For $x = -2$: $-2+5 = C(-2+1)^2 \Rightarrow C = 3$. Comparing coefficients of $x^2$: $0 = A + C \Rightarrow A = -C = -3$. Thus, $f(x) = -\frac{3}{x+1} + \frac{4}{(x+1)^2} + \frac{3}{x+2}$. Differentiating with respect to $x$: $f'(x) = \frac{d}{dx} \left( -3(x+1)^{-1} + 4(x+1)^{-2} + 3(x+2)^{-1} \right)$ $f'(x) = 3(x+1)^{-2} - 8(x+1)^{-3} - 3(x+2)^{-2}$ $f'(x) = \frac{3}{(x+1)^2} - \frac{8}{(x+1)^3} - \frac{3}{(x+2)^2}$.

$\frac{d}{d x}\left(\frac{x+5}{(x+1)^2(x+2)}\right)=$

Similar Questions

If $y = x + \frac{1}{x}$,then

If $y = \cot^{-1}(x^2)$,then $\frac{dy}{dx}$ is equal to

$\frac{d}{dx} \left[ \left( \frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x} \right) \cot 3x \right] =$

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 $y = \cos^{-1}(\tanh x) + \sinh(\sin 6x)$, then $\frac{dy}{dx} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module