What is the derivative of y=(e^{4x^2+1})/8 ?

The derivative of y=(e^{4x^2+1})/8 is e^{4x^2+1}x

What is the first derivative of y=(e^{4x^2+1})/8 ?

The first derivative of y=(e^{4x^2+1})/8 is e^{4x^2+1}x

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derivative of

y = \frac{e ^{4 x^{2} + 1}}{8}

e^{4 x^{2} + 1} x

Solution steps

\frac{d}{d x} (\frac{e ^{4 x^{2} + 1}}{8})

Take the constant out:

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

= \frac{1}{8} \frac{d}{d x} (e^{4 x^{2} + 1})

Apply the chain rule:

e^{4 x^{2} + 1} \frac{d}{d x} (4 x^{2} + 1)

= e^{4 x^{2} + 1} \frac{d}{d x} (4 x^{2} + 1)

\frac{d}{d x} (4 x^{2} + 1) = 8 x

= \frac{1}{8} e^{4 x^{2} + 1} \cdot 8 x

Simplify

\frac{1}{8} e^{4 x^{2} + 1} \cdot 8 x : e^{4 x^{2} + 1} x

= e^{4 x^{2} + 1} x

2 x + 5 x

h (x) = x^{2}

f (x) = \frac{1}{x ^{2} + 3}

\int_{0}^{1} \frac{1}{x ^{0.5} ln ( x )} d x

f (x) = tan (x), at x = π

What is the derivative of y=(e^{4x^2+1})/8 ?
The derivative of y=(e^{4x^2+1})/8 is e^{4x^2+1}x
What is the first derivative of y=(e^{4x^2+1})/8 ?
The first derivative of y=(e^{4x^2+1})/8 is e^{4x^2+1}x