Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation4 Marks
Question
Differentiate the following w. r. t. x.$x^a+x^x+a^x$
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Answer
Let $y=x^a+x^x+a^x$ Here the derivatives of $x^a$ and $a^x$ can be found directly but we can not find the derivative of $x^x$ without the use of logarithm. So the given function is split in to two functions, find their derivatives and then add them. Let $u=x^a+a^x$ and $v=x^x$ $\therefore \quad y=u+v$, where $u$ and $v$ are differentiable functions of $x$. $ \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} $ Now, $u=x^a+a^x$ $ \begin{aligned} & \text { Differentiate w.r.t.x. } \\ & \frac{d u}{d x}=\frac{d}{d x}\left(x^a\right)+\frac{d}{d x}\left(a^x\right) \\ & \frac{d u}{d x}=a x^{-1}+a^x \log a \end{aligned} $ And, $v=x^x$ Taking log of both the sides we get, $ \begin{aligned} & \log v=\log x^x \\ & \log v=x \log x \end{aligned} $ Differentiate w.r. t. $x$. $ \begin{aligned} \frac{d}{d x}(\log v) & =\frac{d}{d x}(x \log x) \\ \frac{1}{v} \frac{d v}{d x} & =x \frac{d}{d x}(\log x)+\log x \frac{d}{d x}(x) \\ \frac{d v}{d x} & =v\left[x \times \frac{1}{x}+\log x(1)\right] \\ \frac{d v}{d x} & =x^x[1+\log x] \end{aligned} $ Substituting (II) and (III) in (I) we get, $ \frac{d y}{d x}=a x^{-1}+a^x \log a+x^x[1+\log x] $
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