Question
Differentiate $3^x+x^3+4 x-5$ with respect to $x$

Answer

Let, $f(x)=3^x+x^3+4 x-5$
$\therefore \quad \frac{d}{d x} f(x)=f^{\prime}(x)$
$\Rightarrow \quad f^{\prime}(x)=\frac{d}{d x}\left(3^x+x^3+4 x-5\right)$
$\begin{array}{l}=3^x \log _e 3+3 x^2+4 \times 1-0 \\ = 3^x \log _e 3+3 x^2+4\end{array}$

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