Question

Differentiate with respect to x
${e}^{ log( { log(x) }^{2} - { logx }^{2} ) }$
This is the question ,please solve it.....​

Answer 1

Answer:-

Let us consider Y as,

$e { }^{log} {}^{(logx {}^{2)} - logx {}^{2} }$

Now, Differentiate both the sides with x² we'll get,

dy/ dx= [ here we have substituted the value of x and y]

$\frac{d \: (e {}^{log(logx) {}^{2} - logx {}^{2} )} }{d \: (log {}^{(} logx {}^{2} )} \: \times \frac{d \: (log(log) {}^{2} - log \: x {}^{2} }{dx }$

Now,

$e {}^{log(logx) {}^{2 - log \ x {}^{2} } } \times ( \frac{d \: (log)(log \: x {}^{2}) }{d \: (logx) {}^{2} } \times \frac{d \: (log \: x) {}^{2} }{d \:( logx)} \times \frac{d \: logx}{dx} - \frac{d \: (log \: x {}^{2}) }{dx {}^{2} } \times \frac{dx {}^{2} }{dx}$

Also, [cancle out log x)

$( \frac{1}{(logx) {}^{2} } \times 2logx \: \times \frac{1}{2x} .2x)$

Also,

$\frac{2}{x}( \frac{1}{logx} - 1)$

Thus, dx/dy ( Taking common log x) will be

$( \frac{2}{x}( \frac{1 - log \: x}{log \: x} )$

Answer 2

Given expression :-

$\sf{e^{log[log(x)^2-logx^2]}}$

OR

$\sf{e^{ln[ln(x^2)-ln^2(x)]}}$

Objective :-

To differentiate the given expression with respect to x.

Solution :-

Let

$\sf{y=e^{ln[ln(x^2)-ln^2(x)]}}$

Solving y by using logarithm laws, we get :-

$\bf{y=2\ ln(x)-ln^2(x)}$

Now, differentiating y with respect to x :-

$\displaystyle \sf{\frac{dy}{dx} =\frac{d}{dx}[2\ ln(x)-ln^2(x)] }$

$\displaystyle \sf{\to \frac{dy}{dx}= 2.\frac{d}{dx}[ln(x)]-\frac{d}{dx}[ln^2(x)] }$

• Derivative of ln(x) is 1/x.

$\displaystyle \sf{\to \frac{dy}{dx}= 2.\frac{1}{x}-2\ ln(x).\frac{d}{dx}[ln(x)] }$

• Derivative of ln(x) is 1/x.

$\displaystyle \sf{\to \frac{dy}{dx}= \frac{2}{x}-2\ ln(x).\frac{1}{x} }\\\\\displaystyle \sf{\to \frac{dy}{dx}= \dfrac{2}{x}-\dfrac{2\ln\left(x\right)}{x}}\\\\\boxed{\displaystyle \bf{\to \frac{dy}{dx}=-\dfrac{2\ln\left(x\right)-2}{x}}}$

______________________

◘ Linear differentiation :-

[ a.b(x) + c.d(x) ]' = a.b' (x) + c.d' (x)

◘ Power rule (Chain rule applied) :-

[ a(x)ⁿ ]' = n.a(x)ⁿ⁻¹ . a'(x)