What is the derivative of f(x)=72x^3(3x^4+7)^5 ?

The derivative of f(x)=72x^3(3x^4+7)^5 is 72(3x^2(3x^4+7)^5+60x^6(3x^4+7)^4)

What is the first derivative of f(x)=72x^3(3x^4+7)^5 ?

The first derivative of f(x)=72x^3(3x^4+7)^5 is 72(3x^2(3x^4+7)^5+60x^6(3x^4+7)^4)

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derivative of

f (x) = 72 x^{3} (3 x^{4} + 7)^{5}

72 (3 x^{2} (3 x^{4} + 7)^{5} + 60 x^{6} (3 x^{4} + 7)^{4})

Solution steps

\frac{d}{d x} (72 x^{3} (3 x^{4} + 7)^{5})

Take the constant out:

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

= 72 \frac{d}{d x} (x^{3} (3 x^{4} + 7)^{5})

Apply the Product Rule:

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

f = x^{3}, g = (3 x^{4} + 7)^{5}

= 72 (\frac{d}{d x} (x^{3}) (3 x^{4} + 7)^{5} + \frac{d}{d x} ((3 x^{4} + 7)^{5}) x^{3})

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

\frac{d}{d x} ((3 x^{4} + 7)^{5}) = 60 x^{3} (3 x^{4} + 7)^{4}

= 72 (3 x^{2} (3 x^{4} + 7)^{5} + 60 x^{3} (3 x^{4} + 7)^{4} x^{3})

60 x^{3} (3 x^{4} + 7)^{4} x^{3} = 60 x^{6} (3 x^{4} + 7)^{4}

= 72 (3 x^{2} (3 x^{4} + 7)^{5} + 60 x^{6} (3 x^{4} + 7)^{4})

d er i v a t i v e f (x) = \frac{x}{4 x + 5}

n = 1 \sum \infty \frac{8}{n ^{3}}

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

\int sec^{3} (π) x d x

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

What is the derivative of f(x)=72x^3(3x^4+7)^5 ?
The derivative of f(x)=72x^3(3x^4+7)^5 is 72(3x^2(3x^4+7)^5+60x^6(3x^4+7)^4)
What is the first derivative of f(x)=72x^3(3x^4+7)^5 ?
The first derivative of f(x)=72x^3(3x^4+7)^5 is 72(3x^2(3x^4+7)^5+60x^6(3x^4+7)^4)