If y = log left[ frac{x + sqrt{x^2 + 25}}{sqr | vedclass

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

Question

(B) Given $y = \log \left[ \frac{x + \sqrt{x^2 + 25}}{\sqrt{x^2 + 25} - x} \right]$.Rationalizing the denominator inside the logarithm:$y = \log \left

Vedclass · Accepted Answer

$\frac{2}{\sqrt{x^2 + 25}}$

Vedclass · Answer

(B) Given $y = \log \left[ \frac{x + \sqrt{x^2 + 25}}{\sqrt{x^2 + 25} - x} \right]$.Rationalizing the denominator inside the logarithm:$y = \log \left[ \frac{(x + \sqrt{x^2 + 25})(x + \sqrt{x^2 + 25})}{(\sqrt{x^2 + 25} - x)(\sqrt{x^2 + 25} + x)} \right]$$y = \log \left[ \frac{(x + \sqrt{x^2 + 25})^2}{(x^2 + 25) - x^2} \right]$$y = \log \left[ \frac{(x + \sqrt{x^2 + 25})^2}{25} \right]$Using logarithmic properties:$y = 2 \log (x + \sqrt{x^2 + 25}) - \log 25$Differentiating with respect to $x$:$\frac{dy}{dx} = 2 \cdot \frac{1}{x + \sqrt{x^2 + 25}} \cdot \frac{d}{dx}(x + \sqrt{x^2 + 25}) - 0$$\frac{dy}{dx} = \frac{2}{x + \sqrt{x^2 + 25}} \cdot \left( 1 + \frac{1}{2\sqrt{x^2 + 25}} \cdot 2x \right)$$\frac{dy}{dx} = \frac{2}{x + \sqrt{x^2 + 25}} \cdot \left( \frac{\sqrt{x^2 + 25} + x}{\sqrt{x^2 + 25}} \right)$$\frac{dy}{dx} = \frac{2}{\sqrt{x^2 + 25}}$

If $y = \log \left[ \frac{x + \sqrt{x^2 + 25}}{\sqrt{x^2 + 25} - x} \right]$,then $\frac{dy}{dx} = \dots$

Similar Questions

Find the derivative: $\frac{d}{dx}(e^{x + 3\log x}) = $

If the slope of the tangent drawn to the curve $y=e^{a+bx^2}$ at the point $P(1,1)$ is $-2$, then the value of $2a-3b$ is

If $f(x) = \log_{x^2}(\log x)$,then at $x = e$,$f'(x)$ has the value

If $y=e^{\log _{e}\left[1+x+x^{2}+\ldots\right]}$,then $\frac{d y}{d x}$ is equal to

Let $f(x)=e^x$, $g(x)=\sin^{-1} x$ and $h(x)=f(g(x))$, then $\frac{h'(x)}{h(x)}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module