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(3) $y = 4^{x}+7ln(x^{2}+1)$
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Question

(3) $y = 4^{x}+7ln(x^{2}+1)$

Explanation:

Step1: Differentiate $4^x$

The derivative of $a^x$ is $a^x\ln a$. So, the derivative of $4^x$ is $4^x\ln 4$.

Step2: Differentiate $7\ln(x^2 + 1)$

Using the chain - rule. Let $u=x^2 + 1$, then $y = 7\ln u$. The derivative of $\ln u$ with respect to $u$ is $\frac{1}{u}$, and the derivative of $u=x^2+1$ with respect to $x$ is $2x$. So the derivative of $7\ln(x^2 + 1)$ is $7\times\frac{1}{x^2 + 1}\times2x=\frac{14x}{x^2 + 1}$.

Step3: Find the derivative of $y$

By the sum - rule of differentiation, if $y = f(x)+g(x)$, then $y^\prime=f^\prime(x)+g^\prime(x)$. So $y^\prime=4^x\ln 4+\frac{14x}{x^2 + 1}$.

Answer:

$y^\prime=4^x\ln 4+\frac{14x}{x^2 + 1}$