What is the derivative of log_{5}(x^6+x) ?

The derivative of log_{5}(x^6+x) is (6x^5+1)/(ln(5)(x^6+x))

What is the first derivative of log_{5}(x^6+x) ?

The first derivative of log_{5}(x^6+x) is (6x^5+1)/(ln(5)(x^6+x))

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\frac{d}{d x} (lo g_{5} (x^{6} + x))

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

Solution steps

\frac{d}{d x} (lo g_{5} (x^{6} + x))

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln ( x ^{6} + x )}{ln ( 5 )})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{ln ( 5 )} \frac{d}{d x} (ln (x^{6} + x))

Apply the chain rule:

\frac{1}{x ^{6} + x} \frac{d}{d x} (x^{6} + x)

= \frac{1}{x ^{6} + x} \frac{d}{d x} (x^{6} + x)

\frac{d}{d x} (x^{6} + x) = 6 x^{5} + 1

= \frac{1}{ln ( 5 )} \cdot \frac{1}{x ^{6} + x} (6 x^{5} + 1)

Simplify

\frac{1}{ln ( 5 )} \cdot \frac{1}{x ^{6} + x} (6 x^{5} + 1) : \frac{6 x ^{5} + 1}{ln ( 5 ) ( x ^{6} + x )}

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

\frac{\partial}{\partial y} (x - 2)

\frac{\partial}{\partial z} (z^{x y^{2}})

\int sec^{2} (3 x) e^{t a n (3 x)} d x

y^{^{'}} - y = x

x \to 0 lim (3 - x^{2})

What is the derivative of log_{5}(x^6+x) ?
The derivative of log_{5}(x^6+x) is (6x^5+1)/(ln(5)(x^6+x))
What is the first derivative of log_{5}(x^6+x) ?
The first derivative of log_{5}(x^6+x) is (6x^5+1)/(ln(5)(x^6+x))