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

The derivative of (e^{4x^{2+1}}/8) is (3e^{4x^3}x^2)/2

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

The first derivative of (e^{4x^{2+1}}/8) is (3e^{4x^3}x^2)/2

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

\frac{3 e ^{4 x^{3}} x ^{2}}{2}

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}) = 12 x^{2}

= \frac{1}{8} e^{4 x^{2 + 1}} \cdot 12 x^{2}

Simplify

\frac{1}{8} e^{4 x^{2 + 1}} \cdot 12 x^{2} : \frac{3 e ^{4 x^{3}} x ^{2}}{2}

= \frac{3 e ^{4 x^{3}} x ^{2}}{2}

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

(x \frac{d y}{d x} + y) (2 (x + x y)) = x y

\frac{x y ^{^{'}}}{y - 2} = y^{2} - y

f (x) = \frac{- 6 x}{x ^{2} + 1}, (1, - 3)

\int π x^{4} d x

What is the derivative of (e^{4x^{2+1}}/8) ?
The derivative of (e^{4x^{2+1}}/8) is (3e^{4x^3}x^2)/2
What is the first derivative of (e^{4x^{2+1}}/8) ?
The first derivative of (e^{4x^{2+1}}/8) is (3e^{4x^3}x^2)/2