Question

$f\left(x\right)={\int }_{e^{2x}}^{e^{3x}}\frac{t}{lnt}dtx>0$.
Find derivative of $f\left(x\right)$ w.r.t. $lnx$ when $x=ln2$

Answer 1

$f\left(x\right)={\int }_{a\left(x\right)}^{b\left(x\right)}g\left(t\right)dt)$

leibnitz formula is$\frac{d\left(f\right(x\left)\right)}{dx}=\frac{d}{dx}\left(b\right(x\left)\right).g\left(b\right(x\left)\right)-\frac{d}{dx}\left(a\right(x\left)\right).g\left(a\right(x\left)\right)$

Here $g\left(t\right)=\frac{t}{\ln t}$

so $\frac{d\left(f\right(x\left)\right)}{dx}=3e^{3x}.\frac{e^{3x}}{3x}-2e^{2x}.\frac{e^{2x}}{2x}=\frac{e^{6x}-e^{4x}}{x}$

derivative w.r.t. $\ln x$ is $\frac{d\left(f\right(x\left)\right)}{d\left(\ln x\right)}$ $=\frac{d\left(f\right(x\left)\right)}{dx}.\frac{dx}{d\left(\ln x\right)}$

$\frac{d\left(f\right(x\left)\right)}{d\left(\ln x\right)}=$ $\frac{e^{6x}-e^{4x}}{x}.x=e^{6x}-e^{4x}$

Now putting the value of $x=\ln 2$ we get

$\frac{d\left(f\right(x\left)\right)}{d\left(\ln x\right)}=$ $e^{6\ln 2}-e^{4\ln 2}=2^6-2^4=48$