Question

differentiate.
y = 6^xcdotlog_4x
\frac{d}{dx}(6^xcdotlog_4x)=square

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u = 6^{x}$ and $v=\log_{4}x$.

Step2: Differentiate $u = 6^{x}$

The derivative of $a^{x}$ with respect to $x$ is $a^{x}\ln a$. So, $u^\prime=\frac{d}{dx}(6^{x})=6^{x}\ln 6$.

Step3: Differentiate $v=\log_{4}x$

We know that $\log_{a}x=\frac{\ln x}{\ln a}$. So, $v = \log_{4}x=\frac{\ln x}{\ln 4}$, and $v^\prime=\frac{d}{dx}(\frac{\ln x}{\ln 4})=\frac{1}{x\ln 4}$.

Step4: Substitute $u$, $u^\prime$, $v$, $v^\prime$ into product - rule

$y^\prime=u^\prime v+uv^\prime=6^{x}\ln 6\cdot\log_{4}x + 6^{x}\cdot\frac{1}{x\ln 4}$.

Answer:

$6^{x}\ln 6\cdot\log_{4}x+\frac{6^{x}}{x\ln 4}$