Solve f(x)=6x^4-11x^3-35^2+34x+24

$$f ( x ) = 6 x ^ { 4 } - 11 x ^ { 3 } - 35 ^ { 2 } + 34 x + 24$$

Evaluate

$$6x^{4}-11x^{3}+34x-1201$$

Solution Steps

Calculate 35 to the power of 2 and get 1225.

$$6x^{4}-11x^{3}-1225+34x+24$$

Add -1225 and 24 to get -1201.

$$6x^{4}-11x^{3}-1201+34x$$

Differentiate w.r.t. x

$$24x^{3}-33x^{2}+34$$

Steps Using Derivative Rule for Sum

Calculate 35 to the power of 2 and get 1225.

$$\frac{\mathrm{d}}{\mathrm{d}x}(6x^{4}-11x^{3}-1225+34x+24)$$

Add -1225 and 24 to get -1201.

$$\frac{\mathrm{d}}{\mathrm{d}x}(6x^{4}-11x^{3}-1201+34x)$$

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}.

$$4\times 6x^{4-1}+3\left(-11\right)x^{3-1}+34x^{1-1}$$

Multiply 4 times 6.

$$24x^{4-1}+3\left(-11\right)x^{3-1}+34x^{1-1}$$

Subtract 1 from 4.

$$24x^{3}+3\left(-11\right)x^{3-1}+34x^{1-1}$$

Multiply 3 times -11.

$$24x^{3}-33x^{3-1}+34x^{1-1}$$

Subtract 1 from 3.

$$24x^{3}-33x^{2}+34x^{1-1}$$

Subtract 1 from 1.

$$24x^{3}-33x^{2}+34x^{0}$$

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

$$24x^{3}-33x^{2}+34\times 1$$

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

$$24x^{3}-33x^{2}+34$$