What is the derivative of a(1+e^{(bx+c)})+d ?

The derivative of a(1+e^{(bx+c)})+d is abe^{bx+c}

What is the first derivative of a(1+e^{(bx+c)})+d ?

The first derivative of a(1+e^{(bx+c)})+d is abe^{bx+c}

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derivative of

a (1 + e^{(b x + c)}) + d

ab e^{b x + c}

Solution steps

\frac{d}{d x} (a (1 + e^{b x + c}) + d)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (a (1 + e^{b x + c})) + \frac{dd}{d x}

\frac{d}{d x} (a (1 + e^{b x + c})) = ab e^{b x + c}

\frac{dd}{d x} = 0

= ab e^{b x + c} + 0

Simplify

= ab e^{b x + c}

\frac{d}{d x} (- 2 x - y^{2} cos (x y^{2}) - 1)

\frac{1}{5} x^{5} - \frac{1}{2 x ^{4}} + 3 x + 9

\int \frac{4 x + 1}{4 x ^{2} + 2 x + 3} d x

2 t \cdot \frac{d y}{d t} + 5 y = t

y = \frac{8 x}{e ^{x}}

What is the derivative of a(1+e^{(bx+c)})+d ?
The derivative of a(1+e^{(bx+c)})+d is abe^{bx+c}
What is the first derivative of a(1+e^{(bx+c)})+d ?
The first derivative of a(1+e^{(bx+c)})+d is abe^{bx+c}