Solve f(x)=3x^4-7x^1-25x^2+63x-18

$$f ( x ) = 3 x ^ { 4 } - 7 x ^ { 1 } - 25 x ^ { 2 } + 63 x - 18$$

Evaluate

$$3x^{4}-25x^{2}+56x-18$$

Solution Steps

Calculate x to the power of 1 and get x.

$$3x^{4}-7x-25x^{2}+63x-18$$

Combine -7x and 63x to get 56x.

$$3x^{4}+56x-25x^{2}-18$$

Differentiate w.r.t. x

$$12x^{3}-50x+56$$

Steps Using Derivative Rule for Sum

Calculate x to the power of 1 and get x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{4}-7x-25x^{2}+63x-18)$$

Combine -7x and 63x to get 56x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{4}+56x-25x^{2}-18)$$

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}.

$$4\times 3x^{4-1}+56x^{1-1}+2\left(-25\right)x^{2-1}$$

Multiply 4 times 3.

$$12x^{4-1}+56x^{1-1}+2\left(-25\right)x^{2-1}$$

Subtract 1 from 4.

$$12x^{3}+56x^{1-1}+2\left(-25\right)x^{2-1}$$

Subtract 1 from 1.

$$12x^{3}+56x^{0}+2\left(-25\right)x^{2-1}$$

Multiply 1 times 56.

$$12x^{3}+56x^{0}-50x^{2-1}$$

Subtract 1 from 2.

$$12x^{3}+56x^{0}-50x^{1}$$

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

$$12x^{3}+56x^{0}-50x$$

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

$$12x^{3}+56\times 1-50x$$

For any term t, t\times 1=t and 1t=t.

$$12x^{3}+56-50x$$