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

The derivative of x^{log_{5}(x)} is (2x^{log_{5}(x)-1}ln(x))/(ln(5))

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

The first derivative of x^{log_{5}(x)} is (2x^{log_{5}(x)-1}ln(x))/(ln(5))

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derivative of

x^{l o g_{5} (x)}

\frac{2 x ^{l o g_{5} (x) - 1} ln ( x )}{ln ( 5 )}

Solution steps

\frac{d}{d x} (x^{l o g_{5} (x)})

Apply exponent rule:

a^{b} = e^{b l n (a)}

x^{l o g_{5} (x)} = e^{l o g_{5} (x) l n (x)}

= \frac{d}{d x} (e^{l o g_{5} (x) l n (x)})

Apply the chain rule:

e^{l o g_{5} (x) l n (x)} \frac{d}{d x} (lo g_{5} (x) ln (x))

= e^{l o g_{5} (x) l n (x)} \frac{d}{d x} (lo g_{5} (x) ln (x))

\frac{d}{d x} (lo g_{5} (x) ln (x)) = \frac{2 ln ( x )}{ln ( 5 ) x}

= e^{l o g_{5} (x) l n (x)} \frac{2 ln ( x )}{ln ( 5 ) x}

Simplify

e^{l o g_{5} (x) l n (x)} \frac{2 ln ( x )}{ln ( 5 ) x} : \frac{2 x ^{l o g_{5} (x) - 1} ln ( x )}{ln ( 5 )}

= \frac{2 x ^{l o g_{5} (x) - 1} ln ( x )}{ln ( 5 )}

\frac{\partial}{\partial x} (\frac{8}{x ^{2} + y ^{2}})

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

t an g e n t x - 1

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

d er i v a t i v e f (x) = x - 3 x

What is the derivative of x^{log_{5}(x)} ?
The derivative of x^{log_{5}(x)} is (2x^{log_{5}(x)-1}ln(x))/(ln(5))
What is the first derivative of x^{log_{5}(x)} ?
The first derivative of x^{log_{5}(x)} is (2x^{log_{5}(x)-1}ln(x))/(ln(5))