Differentiate w.r.t. x the function in Exercise: x^ x^2-3 +(x-3)^ x^2 , for x > 3

Let y = x^ x^ 2-3 +(x-3)^ x^2 Also, let u=x^ x^2-3 and v=(x-3)^ x^2 y=u+v Differentiating both sides with respect to x, we obtain dy dx = du dx + dv dx (1) u=x^ x^2-3 = (x^ x^2-3 ) =(x^2-3) Differentiating with respect to x, we obtain 1 u du dx = . d dx (x^2-3)+(x^2+3) d dx ( ) 1 u du dx = 2x+(x^2-3) 1 x du dx =x^ x^2-3 [ x^2-3 x +2x ] Also, v=(x-3)^ x^2 = (x-3)^ x^2 =x^2 (x-3) Differentiating both sides with respect to x, we obtain 1 v dv dx = (x-3) d dx (x^2)+x^2 d dx [ (x-3) ] 1 v dv dx = (x-3) 2x+x^2 1 x-3 d dx (x-3) dv dx =v [2x (x-3)+ x^2 x-3 1 ] dv dx =(x-3)^ x^2 [ x^2 x-3 +2x (x-3) ] Substituting the expressions of du dx and dv dx in equation (1), we obtain dy dx =x^ x^2-3 [ x^2-3 x +2xlogx ]+(x-3)^x [ x^ x x-3 +2x (x-3) ]

Differentiate w.r.t. x the function in Exercise: x^ x^2-3 +(x-3)^…

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Question

Differentiate w.r.t. x the function in Exercise:
$\text{x}^{\text{x}^2-3}+(\text{x}-3)^{\text{x}^2},$ for x > 3

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Answer

Let y = $\text{x}^{\text{x}^{2-3}}+(\text{x}-3)^{\text{x}^2}$
Also, let $\text{u}=\text{x}^{\text{x}^2-3}\text{ and v}=(\text{x}-3)^{\text{x}^2}$
$\therefore\ \text{y}=\text{u}+\text{v}$
Differentiating both sides with respect to x, we obtain
$\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ \ \dots(1)$
$\text{u}=\text{x}^{\text{x}^2-3}$
$\therefore\ \log\text{u}=\log(\text{x}^{\text{x}^2-3})$
$\log\text{u}=(\text{x}^2-3)\log\text{x}$
Differentiating with respect to x, we obtain
$\frac{1}{\text{u}}\cdot\frac{\text{du}}{\text{dx}}=\log\text{x}.\frac{\text{d}}{\text{dx}}(\text{x}^2-3)+(\text{x}^2+3)\cdot\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log\text{x}\cdot2\text{x}+(\text{x}^2-3)\cdot\frac{1}{\text{x}}$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}^2-3}\cdot\Bigg[\frac{\text{x}^2-3}{\text{x}}+2\text{x}\log\text{x}\Bigg]$
Also,
$\text{v}=(\text{x}-3)^{\text{x}^2}$
$\therefore\ \log\text{v}=\log(\text{x}-3)^{\text{x}^2}$
$\Rightarrow\ \log\text{v}=\text{x}^2\log(\text{x}-3)$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}\cdot\frac{\text{dv}}{\text{dx}}=\log(\text{x}-3)\cdot\frac{\text{d}}{\text{dx}}(\text{x}^2)+\text{x}^2\cdot\frac{\text{d}}{\text{dx}}\big[\log(\text{x}-3)\big]$
$\Rightarrow\ \frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\log(\text{x}-3)\cdot2\text{x}+\text{x}^2\cdot\frac{1}{\text{x}-3}\cdot\frac{\text{d}}{\text{dx}}(\text{x}-3)$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\text{v}\Bigg[2\text{x}\log(\text{x}-3)+\frac{\text{x}^2}{\text{x}-3}\cdot1\Bigg]$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=(\text{x}-3)^{\text{x}^2}\Bigg[\frac{\text{x}^2}{\text{x}-3}+2\text{x}\log(\text{x}-3)\Bigg]$
Substituting the expressions of $\frac{\text{du}}{\text{dx}}\text{ and }\frac{\text{dv}}{\text{dx}}$ in equation (1), we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x}^2-3}\Bigg[\frac{\text{x}^2-3}{\text{x}}+2\text{x}\text{logx}\Bigg]+(\text{x}-3)^\text{x} \Bigg[\frac{\text{x}^{\text{x}}}{\text{x}-3}+2\text{x}\log(\text{x}-3)\Bigg]$

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