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

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

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

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

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derivative of

h (x) = (- 2 x - 3)^{5} (3 x + 1)^{3}

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

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{d}{d x} ((- 2 x - 3)^{5}) = - 10 (- 2 x - 3)^{4}

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

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

Simplify

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

f (x) = x^{4}, at x = 2

\int (\frac{9 + x}{x})^{2} d x

\int - csc (2 x) cot (2 x) d x

(2, 3), (- 4, - 3)

h \to 0 lim (\frac{( 5 + h ) ^{4} - 625}{h})

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