What is the derivative of (e^x-4)/(ln|x|) ?

The derivative of (e^x-4)/(ln|x|) is (e^xxln|x|-e^x+4)/(xln|x|^2)

What is the first derivative of (e^x-4)/(ln|x|) ?

The first derivative of (e^x-4)/(ln|x|) is (e^xxln|x|-e^x+4)/(xln|x|^2)

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derivative of

\frac{e ^{x} - 4}{ln ∣ x ∣}

\frac{e ^{x} x ln ∣ x ∣ - e ^{x} + 4}{x ln ∣ x ∣ ^{2}}

Solution steps

\frac{d}{d x} (\frac{e ^{x} - 4}{ln ∣ x ∣})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( e ^{x} - 4 ) ln ∣ x ∣ - \frac{d}{d x} ( ln ∣ x ∣ ) ( e ^{x} - 4 )}{( ln ∣ x ∣ ) ^{2}}

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

\frac{d}{d x} (ln ∣ x ∣) = \frac{1}{x}

= \frac{e ^{x} ln ∣ x ∣ - \frac{1}{x} ( e ^{x} - 4 )}{( ln ∣ x ∣ ) ^{2}}

Simplify

\frac{e ^{x} ln ∣ x ∣ - \frac{1}{x} ( e ^{x} - 4 )}{ln ∣ x ∣ ^{2}} : \frac{e ^{x} x ln ∣ x ∣ - e ^{x} + 4}{x ln ∣ x ∣ ^{2}}

= \frac{e ^{x} x ln ∣ x ∣ - e ^{x} + 4}{x ln ∣ x ∣ ^{2}}

\int \frac{4}{t t ^{2} + 36} d t

\int x^{2} + 2 x^{4} d x

\frac{d}{d x} (\frac{sin ^{5} ( x )}{5} - \frac{sin ^{7} ( x )}{7})

\frac{d}{d x} (e^{x} - cos (x))

y = (2 x + 1) (x + 5)^{3}

What is the derivative of (e^x-4)/(ln|x|) ?
The derivative of (e^x-4)/(ln|x|) is (e^xxln|x|-e^x+4)/(xln|x|^2)
What is the first derivative of (e^x-4)/(ln|x|) ?
The first derivative of (e^x-4)/(ln|x|) is (e^xxln|x|-e^x+4)/(xln|x|^2)