text{If } frac{d}{dx} left( frac{x^2+1}{(x^2+ | vedclass

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

Question

(A) Let $y = \frac{x^2+1}{(x^2+5)(x^2+9)}$.Taking the natural logarithm on both sides:$\log y = \log(x^2+1) - \log(x^2+5) - \log(x^2+9)$.Differentiati

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) Let $y = \frac{x^2+1}{(x^2+5)(x^2+9)}$.Taking the natural logarithm on both sides:$\log y = \log(x^2+1) - \log(x^2+5) - \log(x^2+9)$.Differentiating with respect to $x$:$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{2x}{x^2+1} - \frac{2x}{x^2+5} - \frac{2x}{x^2+9}$.$\frac{dy}{dx} = y \cdot 2x \left[ \frac{1}{x^2+1} - \frac{1}{x^2+5} - \frac{1}{x^2+9} \right]$.Substituting $y = \frac{x^2+1}{(x^2+5)(x^2+9)}$:$\frac{dy}{dx} = \frac{2x(x^2+1)}{(x^2+5)(x^2+9)} \left[ \frac{1}{x^2+1} - \frac{1}{x^2+5} - \frac{1}{x^2+9} \right]$.Comparing this with the given expression,we get:$f(x) = x^2+1, g(x) = x^2+5, h(x) = x^2+9$.Now,calculate $2h(x) - f(x) - g(x)$:$2(x^2+9) - (x^2+1) - (x^2+5) = 2x^2 + 18 - x^2 - 1 - x^2 - 5 = 12$.

$\text{If } \frac{d}{dx} \left( \frac{x^2+1}{(x^2+5)(x^2+9)} \right) = \frac{2x(x^2+1)}{(x^2+5)(x^2+9)} \left[ \frac{1}{f(x)} - \frac{1}{g(x)} - \frac{1}{h(x)} \right], \text{ then } 2h(x) - f(x) - g(x) = $

Similar Questions

If $h(x) = x^{x^x}$, then at $x = 1$, $\frac{h^{\prime}(x)}{h(x)}$ is equal to

If $y = x^{\ln x}$,then $dy/dx$ equals :-

If $x^{p} y^{q}=(x+y)^{p+q}$,then $\frac{d y}{d x}$ is equal to

The rate of change of $x^{\sin x}$ with respect to $(\sin x)^{x}$ is

If $y = 2x^{3x}$,then $\frac{dy}{dx}$ at $x = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module