#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 18 sub question (viii) maths textbook solution

Answer:  $\left(\frac{x^{2}-3}{x}+2 x \log x\right) x^{x^{2}-3}+\left(\frac{x^{2}}{x-3}+2 x \log (x-3)\right)(x-3)^{x^{2}}$

Hint: Diff by $x^{n-3}$

Given: $y=x^{x^{2}-3}+(x-3)^{x^{2}}$

Solution:  $y=u+v$

$u=x^{x^{2}-3}$

\begin{aligned} &\log u=\log x^{x^{2}-3} \\\\ &\log u=\left(x^{2}-3\right) \log x \end{aligned}

$\frac{1}{u} \frac{d u}{d x}=\left(x^{2}-3\right) \cdot \frac{1}{x}+\log x(2 x)$

$\frac{d u}{d x}=\left(\frac{x^{2}-3}{x}+2 x \log x\right) \cdot x^{x^{2}-3}$

Now $v=(x-3)^{x^{2}}$

Take log on both sides

\begin{aligned} &\log v=\log (x-3)^{x^{2}} \\\\ &\frac{1}{v} \frac{d v}{d x}=x^{2} \log (x-3) \end{aligned}

$\frac{d v}{d x}=\left[\frac{x^{2}}{x-3}+2 x \log (x-3)\right](x-3)^{x^{2}}$

$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

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