What is the derivative of (6x-7^3(8x^2+9)^2) ?

The derivative of (6x-7^3(8x^2+9)^2) is 18(6x-7)^2(8x^2+9)^2+32x(8x^2+9)(6x-7)^3

What is the first derivative of (6x-7^3(8x^2+9)^2) ?

The first derivative of (6x-7^3(8x^2+9)^2) is 18(6x-7)^2(8x^2+9)^2+32x(8x^2+9)(6x-7)^3

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\frac{d}{d x} ((6 x - 7)^{3} (8 x^{2} + 9)^{2})

18 (6 x - 7)^{2} (8 x^{2} + 9)^{2} + 32 x (8 x^{2} + 9) (6 x - 7)^{3}

Solution steps

\frac{d}{d x} ((6 x - 7)^{3} (8 x^{2} + 9)^{2})

Apply the Product Rule:

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

f = (6 x - 7)^{3}, g = (8 x^{2} + 9)^{2}

= \frac{d}{d x} ((6 x - 7)^{3}) (8 x^{2} + 9)^{2} + \frac{d}{d x} ((8 x^{2} + 9)^{2}) (6 x - 7)^{3}

\frac{d}{d x} ((6 x - 7)^{3}) = 18 (6 x - 7)^{2}

\frac{d}{d x} ((8 x^{2} + 9)^{2}) = 32 x (8 x^{2} + 9)

= 18 (6 x - 7)^{2} (8 x^{2} + 9)^{2} + 32 x (8 x^{2} + 9) (6 x - 7)^{3}

\int_{0}^{5} (5 - x) d x

x y d y = y d x

\frac{4}{x}, \frac{4}{x ^{2}}, x = 7

9 y^{^{''}} - y = x e^{\frac{x}{3}}

\frac{d y}{d x}, x + y = x y

What is the derivative of (6x-7^3(8x^2+9)^2) ?
The derivative of (6x-7^3(8x^2+9)^2) is 18(6x-7)^2(8x^2+9)^2+32x(8x^2+9)(6x-7)^3
What is the first derivative of (6x-7^3(8x^2+9)^2) ?
The first derivative of (6x-7^3(8x^2+9)^2) is 18(6x-7)^2(8x^2+9)^2+32x(8x^2+9)(6x-7)^3