What is the derivative of ln(e^{ax+b}) ?

The derivative of ln(e^{ax+b}) is a

What is the first derivative of ln(e^{ax+b}) ?

The first derivative of ln(e^{ax+b}) is a

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\frac{d}{d x} (ln (e^{a x + b}))

a

Solution steps

\frac{d}{d x} (ln (e^{a x + b}))

Simplify

ln (e^{a x + b}) : a x + b

= \frac{d}{d x} (a x + b)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (a x) + \frac{d b}{d x}

\frac{d}{d x} (a x) = a

\frac{d b}{d x} = 0

= a + 0

Simplify

= a

\int \frac{8}{x ^{7}} - \frac{6}{x ^{4}} + 3 x d x

y = 0.4 x^{1.2}

y = x (x - 8), (16, 32)

\int_{0}^{1} x^{2} - 2 x + 1 d x

f (x) = 3 x + 1, (5, 4)

What is the derivative of ln(e^{ax+b}) ?
The derivative of ln(e^{ax+b}) is a
What is the first derivative of ln(e^{ax+b}) ?
The first derivative of ln(e^{ax+b}) is a