What is the derivative of y=log_{5}(x^4-7) ?

The derivative of y=log_{5}(x^4-7) is (4x^3)/(ln(5)(x^4-7))

What is the first derivative of y=log_{5}(x^4-7) ?

The first derivative of y=log_{5}(x^4-7) is (4x^3)/(ln(5)(x^4-7))

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derivative of

y = lo g_{5} (x^{4} - 7)

\frac{4 x ^{3}}{ln ( 5 ) ( x ^{4} - 7 )}

Solution steps

\frac{d}{d x} (lo g_{5} (x^{4} - 7))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( x ^{4} - 7 )}{ln ( 5 )})

Take the constant out:

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

= \frac{1}{ln ( 5 )} \frac{d}{d x} (ln (x^{4} - 7))

Apply the chain rule:

\frac{1}{x ^{4} - 7} \frac{d}{d x} (x^{4} - 7)

= \frac{1}{x ^{4} - 7} \frac{d}{d x} (x^{4} - 7)

\frac{d}{d x} (x^{4} - 7) = 4 x^{3}

= \frac{1}{ln ( 5 )} \cdot \frac{1}{x ^{4} - 7} \cdot 4 x^{3}

Simplify

\frac{1}{ln ( 5 )} \cdot \frac{1}{x ^{4} - 7} \cdot 4 x^{3} : \frac{4 x ^{3}}{ln ( 5 ) ( x ^{4} - 7 )}

= \frac{4 x ^{3}}{ln ( 5 ) ( x ^{4} - 7 )}

y = x^{3} - 9 x, (1, - 8)

\frac{\partial}{\partial y} (- e^{- x y^{2}} \cdot y^{2})

\frac{\partial}{\partial x} (e^{s i n (x^{2})})

\int \frac{5 arcsin ( x )}{x ^{2}} d x

x^{3} y^{^{'}} + 2 x^{2} y = - 3 cos (3 x) y^{- \frac{1}{2}}

What is the derivative of y=log_{5}(x^4-7) ?
The derivative of y=log_{5}(x^4-7) is (4x^3)/(ln(5)(x^4-7))
What is the first derivative of y=log_{5}(x^4-7) ?
The first derivative of y=log_{5}(x^4-7) is (4x^3)/(ln(5)(x^4-7))