Solve 9^4-8a^3+24a^2-32a+16

$$9 ^ { 4 } - 8 a ^ { 3 } + 24 a ^ { 2 } - 32 a + 16$$

Evaluate

$$6577-32a+24a^{2}-8a^{3}$$

Solution Steps

Calculate 9 to the power of 4 and get 6561.

$$6561-8a^{3}+24a^{2}-32a+16$$

Add 6561 and 16 to get 6577.

$$6577-8a^{3}+24a^{2}-32a$$

Differentiate w.r.t. a

$$-24a^{2}+48a-32$$

Steps Using Derivative Rule for Sum

Calculate 9 to the power of 4 and get 6561.

$$\frac{\mathrm{d}}{\mathrm{d}a}(6561-8a^{3}+24a^{2}-32a+16)$$

Add 6561 and 16 to get 6577.

$$\frac{\mathrm{d}}{\mathrm{d}a}(6577-8a^{3}+24a^{2}-32a)$$

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}.

$$3\left(-8\right)a^{3-1}+2\times 24a^{2-1}-32a^{1-1}$$

Multiply 3 times -8.

$$-24a^{3-1}+2\times 24a^{2-1}-32a^{1-1}$$

Subtract 1 from 3.

$$-24a^{2}+2\times 24a^{2-1}-32a^{1-1}$$

Multiply 2 times 24.

$$-24a^{2}+48a^{2-1}-32a^{1-1}$$

Subtract 1 from 2.

$$-24a^{2}+48a^{1}-32a^{1-1}$$

Subtract 1 from 1.

$$-24a^{2}+48a^{1}-32a^{0}$$

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

$$-24a^{2}+48a-32a^{0}$$

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

$$-24a^{2}+48a-32$$