Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$\log _{x} 3$

Solution:

Let $y=\log _{x} 3$

Recall that $\log _{a} b=\frac{\log _{b}}{\log a}$.

$\Rightarrow \log _{x} 3=\frac{\log 3}{\log x}$

On differentiating $y$ with respect to $x$, we get

$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\log 3}{\log x}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\log 3 \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\log \mathrm{x}}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\log 3 \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})^{-1}$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\log 3\left[-1 \times(\log \mathrm{x})^{-1-1}\right] \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})[$ using chain rule $]$

$\Rightarrow \frac{d y}{d x}=-\log 3(\log x)^{-2} \frac{d}{d x}(\log x)$

We have $\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}$

$\Rightarrow \frac{d y}{d x}=-\log 3(\log x)^{-2} \times \frac{1}{x}$

$\Rightarrow \frac{d y}{d x}=-\frac{1}{x} \frac{\log 3}{(\log x)^{2}}$

$\Rightarrow \frac{d y}{d x}=-\frac{1}{x} \frac{\log 3}{(\log x)^{2}} \times \frac{\log 3}{\log 3}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{\mathrm{x} \log 3} \frac{(\log 3)^{2}}{(\log \mathrm{x})^{2}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{\mathrm{x} \log 3}\left(\frac{\log 3}{\log \mathrm{x}}\right)^{2}$

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

Thus, $\frac{\mathrm{d}}{\mathrm{dx}}\left(\log _{\mathrm{x}} 3\right)=-\frac{1}{x \log 3\left(\log _{3} \mathrm{x}\right)^{2}}$

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