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differentiate. y = log₆(x⁵ + x) \\frac{d}{dx}log₆(x⁵ + x) = \\square

Question

differentiate.
y = log₆(x⁵ + x)
\frac{d}{dx}log₆(x⁵ + x) = \square

Explanation:

Step1: Use the formula for the derivative of $\log_a u$

The formula is $\frac{d}{dx}\log_a u=\frac{1}{u\ln a}\cdot\frac{du}{dx}$. Here $a = 6$ and $u=x^{5}+x$.
So, $\frac{d}{dx}\log_6(x^{5}+x)=\frac{1}{(x^{5}+x)\ln 6}\cdot\frac{d}{dx}(x^{5}+x)$

Step2: Differentiate $u=x^{5}+x$

Using the power rule $\frac{d}{dx}x^{n}=nx^{n - 1}$, we have $\frac{d}{dx}(x^{5}+x)=5x^{4}+1$

Answer:

$\frac{5x^{4}+1}{(x^{5}+x)\ln 6}$