What is the derivative of 48x(4x^2+9)^5 ?

The derivative of 48x(4x^2+9)^5 is 48((4x^2+9)^5+40x^2(4x^2+9)^4)

What is the first derivative of 48x(4x^2+9)^5 ?

The first derivative of 48x(4x^2+9)^5 is 48((4x^2+9)^5+40x^2(4x^2+9)^4)

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derivative of

48 x (4 x^{2} + 9)^{5}

48 ((4 x^{2} + 9)^{5} + 40 x^{2} (4 x^{2} + 9)^{4})

Solution steps

\frac{d}{d x} (48 x (4 x^{2} + 9)^{5})

Take the constant out:

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

= 48 \frac{d}{d x} (x (4 x^{2} + 9)^{5})

Apply the Product Rule:

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

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

= 48 (\frac{d x}{d x} (4 x^{2} + 9)^{5} + \frac{d}{d x} ((4 x^{2} + 9)^{5}) x)

\frac{d x}{d x} = 1

\frac{d}{d x} ((4 x^{2} + 9)^{5}) = 40 x (4 x^{2} + 9)^{4}

= 48 (1 \cdot (4 x^{2} + 9)^{5} + 40 x (4 x^{2} + 9)^{4} x)

Simplify

48 (1 \cdot (4 x^{2} + 9)^{5} + 40 x (4 x^{2} + 9)^{4} x) : 48 ((4 x^{2} + 9)^{5} + 40 x^{2} (4 x^{2} + 9)^{4})

= 48 ((4 x^{2} + 9)^{5} + 40 x^{2} (4 x^{2} + 9)^{4})

(t^{2} - 4 t) (t^{2} + 4 t)

\int \frac{1}{tan ( x )} d x

(x + 1)^{2} y^{^{'}} + 5 (x + 1) y = 16

i = 1 \sum \infty \frac{1}{i ^{2}}

\int e^{- 2 a x} d x

What is the derivative of 48x(4x^2+9)^5 ?
The derivative of 48x(4x^2+9)^5 is 48((4x^2+9)^5+40x^2(4x^2+9)^4)
What is the first derivative of 48x(4x^2+9)^5 ?
The first derivative of 48x(4x^2+9)^5 is 48((4x^2+9)^5+40x^2(4x^2+9)^4)