What is the derivative of (3x-1^4(-2x+9)^5) ?

The derivative of (3x-1^4(-2x+9)^5) is 12(3x-1)^3(-2x+9)^5-10(-2x+9)^4(3x-1)^4

What is the first derivative of (3x-1^4(-2x+9)^5) ?

The first derivative of (3x-1^4(-2x+9)^5) is 12(3x-1)^3(-2x+9)^5-10(-2x+9)^4(3x-1)^4

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\frac{d}{d x} ((3 x - 1)^{4} (- 2 x + 9)^{5})

12 (3 x - 1)^{3} (- 2 x + 9)^{5} - 10 (- 2 x + 9)^{4} (3 x - 1)^{4}

Solution steps

\frac{d}{d x} ((3 x - 1)^{4} (- 2 x + 9)^{5})

Apply the Product Rule:

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

f = (3 x - 1)^{4}, g = (- 2 x + 9)^{5}

= \frac{d}{d x} ((3 x - 1)^{4}) (- 2 x + 9)^{5} + \frac{d}{d x} ((- 2 x + 9)^{5}) (3 x - 1)^{4}

\frac{d}{d x} ((3 x - 1)^{4}) = 12 (3 x - 1)^{3}

\frac{d}{d x} ((- 2 x + 9)^{5}) = - 10 (- 2 x + 9)^{4}

= 12 (3 x - 1)^{3} (- 2 x + 9)^{5} + (- 10 (- 2 x + 9)^{4}) (3 x - 1)^{4}

Simplify

= 12 (3 x - 1)^{3} (- 2 x + 9)^{5} - 10 (- 2 x + 9)^{4} (3 x - 1)^{4}

\frac{d}{d p} (\frac{400}{p ^{2}})

\int (9 x^{2} - 6 x + 5) d x

f (x) = \frac{x}{x + \frac{c}{x}}

y^{^{'}} = (cos^{2} (x)) (cos^{2} (2 y))

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

What is the derivative of (3x-1^4(-2x+9)^5) ?
The derivative of (3x-1^4(-2x+9)^5) is 12(3x-1)^3(-2x+9)^5-10(-2x+9)^4(3x-1)^4
What is the first derivative of (3x-1^4(-2x+9)^5) ?
The first derivative of (3x-1^4(-2x+9)^5) is 12(3x-1)^3(-2x+9)^5-10(-2x+9)^4(3x-1)^4