What is the derivative of f(x)=-5x(2x-3)^4 ?

The derivative of f(x)=-5x(2x-3)^4 is -5((2x-3)^4+8x(2x-3)^3)

What is the first derivative of f(x)=-5x(2x-3)^4 ?

The first derivative of f(x)=-5x(2x-3)^4 is -5((2x-3)^4+8x(2x-3)^3)

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derivative of

f (x) = - 5 x (2 x - 3)^{4}

- 5 ((2 x - 3)^{4} + 8 x (2 x - 3)^{3})

Solution steps

\frac{d}{d x} (- 5 x (2 x - 3)^{4})

Take the constant out:

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

= - 5 \frac{d}{d x} (x (2 x - 3)^{4})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = (2 x - 3)^{4}

= - 5 (\frac{d x}{d x} (2 x - 3)^{4} + \frac{d}{d x} ((2 x - 3)^{4}) x)

\frac{d x}{d x} = 1

\frac{d}{d x} ((2 x - 3)^{4}) = 8 (2 x - 3)^{3}

= - 5 (1 \cdot (2 x - 3)^{4} + 8 (2 x - 3)^{3} x)

Simplify

= - 5 ((2 x - 3)^{4} + 8 x (2 x - 3)^{3})

y^{^{''}} - 2 y^{^{'}} + 5 y = 5, y (0) = 2, y^{^{'}} (0) = 1

\int \frac{( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )} d x

f (x) = (\frac{x - 3}{x + 1})^{2}, at x = 0

\int y e^{3 y} d y

n = 0 \sum \infty \frac{1}{3 + 2 n}

What is the derivative of f(x)=-5x(2x-3)^4 ?
The derivative of f(x)=-5x(2x-3)^4 is -5((2x-3)^4+8x(2x-3)^3)
What is the first derivative of f(x)=-5x(2x-3)^4 ?
The first derivative of f(x)=-5x(2x-3)^4 is -5((2x-3)^4+8x(2x-3)^3)