Solve d/dxleft((-4x)left(x^2+4right)^2right)

$$\frac{d}{d x } \left((-4x) { \left( { x }^{ 2 } +4 \right) }^{ 2 } \right)$$

Evaluate

$$-20x^{4}-96x^{2}-64$$

Steps Using Derivative Rule for Sum

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{2}+4\right)^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-4x\left(\left(x^{2}\right)^{2}+8x^{2}+16\right))$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-4x\left(x^{4}+8x^{2}+16\right))$$

Use the distributive property to multiply -4x by x^{4}+8x^{2}+16.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-4x^{5}-32x^{3}-64x)$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$5\left(-4\right)x^{5-1}+3\left(-32\right)x^{3-1}-64x^{1-1}$$

Multiply 5 times -4.

$$-20x^{5-1}+3\left(-32\right)x^{3-1}-64x^{1-1}$$

Subtract 1 from 5.

$$-20x^{4}+3\left(-32\right)x^{3-1}-64x^{1-1}$$

Multiply 3 times -32.

$$-20x^{4}-96x^{3-1}-64x^{1-1}$$

Subtract 1 from 3.

$$-20x^{4}-96x^{2}-64x^{1-1}$$

Subtract 1 from 1.

$$-20x^{4}-96x^{2}-64x^{0}$$

For any term t except 0, t^{0}=1.

$$-20x^{4}-96x^{2}-64$$

Differentiate w.r.t. x

$$-80x^{3}-192x$$