જો y = sqrt {frac{{1 + x}}{{1 - x}}} , હોય તો | vedclass

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

Question

(B) આપેલ છે $y = \sqrt {\frac{{1 + x}}{{1 - x}}} .$બંને બાજુ પ્રાકૃતિક લઘુગણક લેતા:$\ln y = \frac{1}{2} \ln(1 + x) - \frac{1}{2} \ln(1 - x).$બંને બાજુ

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) આપેલ છે $y = \sqrt {\frac{{1 + x}}{{1 - x}}} .$બંને બાજુ પ્રાકૃતિક લઘુગણક લેતા:$\ln y = \frac{1}{2} \ln(1 + x) - \frac{1}{2} \ln(1 - x).$બંને બાજુ $x$ ની સાપેક્ષમાં વિકલન કરતા:$\frac{1}{y} \frac{{dy}}{{dx}} = \frac{1}{2(1 + x)} - \frac{1}{2} \cdot \frac{1}{(1 - x)} \cdot (-1)$$\frac{1}{y} \frac{{dy}}{{dx}} = \frac{1}{2(1 + x)} + \frac{1}{2(1 - x)}$$\frac{1}{y} \frac{{dy}}{{dx}} = \frac{1}{2} \left[ \frac{(1 - x) + (1 + x)}{(1 + x)(1 - x)} \right]$$\frac{1}{y} \frac{{dy}}{{dx}} = \frac{1}{2} \left[ \frac{2}{1 - {x^2}} \right] = \frac{1}{1 - {x^2}}$$\frac{{dy}}{{dx}} = \frac{y}{1 - {x^2}} = \frac{1}{1 - {x^2}} \cdot \sqrt {\frac{{1 + x}}{{1 - x}}} $$\frac{{dy}}{{dx}} = \frac{1}{{(1 - x)(1 + x)}} \cdot \frac{{{{(1 + x)}^{1/2}}}}{{{{(1 - x)}^{1/2}}}} = \frac{{{{(1 + x)}^{1/2}}}}{{{{(1 + x)}^1}{{(1 - x)}^{3/2}}}} = \frac{1}{{{{(1 + x)}^{1/2}}{{(1 - x)}^{3/2}}}}.$

જો $y = \sqrt {\frac{{1 + x}}{{1 - x}}} ,$ હોય તો $\frac{{dy}}{{dx}} = $

Similar Questions

જો $y = \frac{2(x - \sin x)^{3/2}}{\sqrt{x}}$ હોય,તો $\frac{dy}{dx} = $

જો $y = (x \log x)^{\log \log x}$ હોય,તો $\frac{dy}{dx} = $

જો $y=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}$ હોય, તો $\frac{d y}{d x}=$

જો $y = e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$ હોય,તો $\frac{dy}{dx} = $

$\frac{d}{dx}\{(\sin x)^x\} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module