Question

Differentiate 3^x w.r.t. log_x3.

Answer 1

Let u = 3^x and v = log_x3.
Then we want to find ${\displaystyle \frac{\text{du}}{\text{dv}}}$.
Differentiating u and v w.r.t. x, we get
${\displaystyle \frac{\text{du}}{\text{dv}}=\frac{\text{d}}{\text{dx}}\left(3^x\right)}$
= 3^x.log3
and
${\displaystyle \frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\left({\log }_{x}3\right)}$

= ${\displaystyle \frac{\text{d}}{\text{dx}}\left(\frac{\log 3}{\log x}\right)}$

= ${\displaystyle \log 3.\frac{\text{d}}{\text{dx}}{\left(\log x\right)}^{-1}}$

= ${\displaystyle \left(\log 3\right)(-1){\left(\log x\right)}^{-2}.\frac{\text{d}}{\text{dx}}\left(\log x\right)}$

= ${\displaystyle \frac{-\log 3}{{\left(\log x\right)}^2}\times \frac{1}{x}}$

= ${\displaystyle \frac{-\log 3}{x{\left(\log x\right)}^2}}$

∴ ${\displaystyle \frac{\text{du}}{\text{dv}}=\frac{\left(\frac{\text{du}}{\text{dx}}\right)}{\left(\frac{\text{dv}}{\text{dx}}\right)}}$

= ${\displaystyle \frac{3^x.\log 3}{\left[\frac{-\log 3}{x{\left(\log x\right)}^2}\right]}}$
= – x(log x)².3^x.