Solve {18}^41-frac{35{5}}+12x+5x^2=

$${ 18 }^{ 41- \frac{ 35 }{ 5 } } +12x+5 { x }^{ 2 } =$$

Evaluate

$$5x^{2}+12x+4778209346999261057189801002565550445953024$$

Solution Steps

Divide 35 by 5 to get 7.

$$18^{41-7}+12x+5x^{2}$$

Subtract 7 from 41 to get 34.

$$18^{34}+12x+5x^{2}$$

Calculate 18 to the power of 34 and get 4778209346999261057189801002565550445953024.

$$4778209346999261057189801002565550445953024+12x+5x^{2}$$

Differentiate w.r.t. x

$$10x+12$$

Steps Using Derivative Rule for Sum

Divide 35 by 5 to get 7.

$$\frac{\mathrm{d}}{\mathrm{d}x}(18^{41-7}+12x+5x^{2})$$

Subtract 7 from 41 to get 34.

$$\frac{\mathrm{d}}{\mathrm{d}x}(18^{34}+12x+5x^{2})$$

Calculate 18 to the power of 34 and get 4778209346999261057189801002565550445953024.

$$\frac{\mathrm{d}}{\mathrm{d}x}(4778209346999261057189801002565550445953024+12x+5x^{2})$$

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}.

$$12x^{1-1}+2\times 5x^{2-1}$$

Subtract 1 from 1.

$$12x^{0}+2\times 5x^{2-1}$$

Multiply 2 times 5.

$$12x^{0}+10x^{2-1}$$

Subtract 1 from 2.

$$12x^{0}+10x^{1}$$

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

$$12x^{0}+10x$$

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

$$12\times 1+10x$$

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

$$12+10x$$