Continuity and Differentiability — Maths STD 12 Science — Question

$u + v + w = {a^b}$

Therefore, $\frac{{du}}{{dx}} + \frac{{dw}}{{dx}} + \frac{{dv}}{{dx}} = 0$ ...(1)

$u = {y^x}$

Taking log both sides, we get

$\log u = \log {y^x}$

$\log u = x.\log y$

Differentiate both sides w.r.t. to x

$\frac{1}{u}.\frac{{du}}{{dx}} = x.\frac{1}{y}.\frac{{dy}}{{dx}} + \log y.1$

$\frac{{du}}{{dx}} = u\left[ {\frac{x}{y}.\frac{{dy}}{{dx}} + \log y} \right]$

$\frac{{du}}{{dx}} = {y^x}\left( {\frac{x}{y}.\frac{{dy}}{{dx}} + \log y} \right)$...(2)

$v = {x^y}$

Taking log both sides, we get

$\log v = \log {x^y}$

$\log v = y.\log x$

$\frac{1}{v}.\frac{{dv}}{{dx}} = y.\frac{1}{x} + \log x.\frac{{dy}}{{dx}}$

$\frac{{dv}}{{dx}} = v\left[ {\frac{y}{x} + \log x.\frac{{dy}}{{dx}}} \right]$

$\frac{{dv}}{{dx}} = {x^y}\left[ {\frac{y}{x} + \log x.\frac{{dy}}{{dx}}} \right]$...(3)

$w = {x^x}$

Taking log both sides, we get

$\log w = \log {x^x}$

$\log w = x\log x$

$\frac{1}{w}.\frac{{dw}}{{dx}} = x.\frac{1}{x} + \log x.1$

$\frac{1}{w}.\frac{{dw}}{{dx}} = 1 + \log x$

$\frac{{dw}}{{dx}} = w\left( {1 + \log x} \right)$

$\frac{{dw}}{{dx}} = {x^x}\left( {1 + \log x} \right)$... (4)

$\frac{{dy}}{{dx}} = \frac{{ - {x^x}\left( {1 + \log x} \right) - y.{x^{y - 1}} - {y^x}\log y}}{{x.{y^{x - 1}} + {x^y}\log x}}$ (by putting(2), (3) and (4) in (1))