What is the derivative of (-3x+2)^4(-2x+3)^3 ?

The derivative of (-3x+2)^4(-2x+3)^3 is -12(-3x+2)^3(-2x+3)^3-6(-2x+3)^2(-3x+2)^4

What is the first derivative of (-3x+2)^4(-2x+3)^3 ?

The first derivative of (-3x+2)^4(-2x+3)^3 is -12(-3x+2)^3(-2x+3)^3-6(-2x+3)^2(-3x+2)^4

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derivative of

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

- 12 (- 3 x + 2)^{3} (- 2 x + 3)^{3} - 6 (- 2 x + 3)^{2} (- 3 x + 2)^{4}

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{d}{d x} ((- 3 x + 2)^{4}) = - 12 (- 3 x + 2)^{3}

\frac{d}{d x} ((- 2 x + 3)^{3}) = - 6 (- 2 x + 3)^{2}

= (- 12 (- 3 x + 2)^{3}) (- 2 x + 3)^{3} + (- 6 (- 2 x + 3)^{2}) (- 3 x + 2)^{4}

Simplify

= - 12 (- 3 x + 2)^{3} (- 2 x + 3)^{3} - 6 (- 2 x + 3)^{2} (- 3 x + 2)^{4}

x \to 0 lim (\frac{2 tan ( \frac{x}{2} )}{x})

\int tan^{5} (6 x) sec^{2} (6 x) d x

\frac{d}{d x} (\frac{1}{9 x ^{\frac{8}{9}}})

f (x) = ln (5) x^{4}

\frac{d}{d x} (\frac{2}{( x + 2 ) ^{3}})

What is the derivative of (-3x+2)^4(-2x+3)^3 ?
The derivative of (-3x+2)^4(-2x+3)^3 is -12(-3x+2)^3(-2x+3)^3-6(-2x+3)^2(-3x+2)^4
What is the first derivative of (-3x+2)^4(-2x+3)^3 ?
The first derivative of (-3x+2)^4(-2x+3)^3 is -12(-3x+2)^3(-2x+3)^3-6(-2x+3)^2(-3x+2)^4