What is the derivative of y=-8^{5x^2-4} ?

The derivative of y=-8^{5x^2-4} is -15ln(2)*2^{15x^2-11}x

What is the first derivative of y=-8^{5x^2-4} ?

The first derivative of y=-8^{5x^2-4} is -15ln(2)*2^{15x^2-11}x

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derivative of

y = - 8^{5 x^{2} - 4}

- 15 ln (2) \cdot 2^{15 x^{2} - 11} x

Solution steps

\frac{d}{d x} (- 8^{5 x^{2} - 4})

Take the constant out:

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

= - \frac{d}{d x} (8^{5 x^{2} - 4})

Apply exponent rule:

a^{b} = e^{b l n (a)}

8^{5 x^{2} - 4} = e^{(5 x^{2} - 4) l n (8)}

= - \frac{d}{d x} (e^{(5 x^{2} - 4) l n (8)})

Apply the chain rule:

e^{(5 x^{2} - 4) l n (8)} \frac{d}{d x} ((5 x^{2} - 4) ln (8))

= e^{(5 x^{2} - 4) l n (8)} \frac{d}{d x} ((5 x^{2} - 4) ln (8))

\frac{d}{d x} ((5 x^{2} - 4) ln (8)) = 30 ln (2) x

= - e^{(5 x^{2} - 4) l n (8)} \cdot 30 ln (2) x

Simplify

- e^{(5 x^{2} - 4) l n (8)} \cdot 30 ln (2) x : - 15 ln (2) \cdot 2^{15 x^{2} - 11} x

= - 15 ln (2) \cdot 2^{15 x^{2} - 11} x

x \to - 1 lim (3 x^{4} - 2 x^{3} + 4 x)

\frac{\partial}{\partial x} (ln (1 + z) + x y^{2} + z)

in v erse l a pl a ce \frac{1}{2 s ^{3} + 10 s ^{2} + 12 s}

d er i v a t i v e y = x^{4} cos (4 x^{2})

a re a y = 17 x, y = x^{2}

What is the derivative of y=-8^{5x^2-4} ?
The derivative of y=-8^{5x^2-4} is -15ln(2)*2^{15x^2-11}x
What is the first derivative of y=-8^{5x^2-4} ?
The first derivative of y=-8^{5x^2-4} is -15ln(2)*2^{15x^2-11}x