Solve (2r^2+sqrt{5})(2r^2-sqrt{5})

$$( 2 r ^ { 2 } + \sqrt { 5 } ) ( 2 r ^ { 2 } - \sqrt { 5 } )$$

Evaluate

$$4r^{4}-5$$

Solution Steps

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(2r^{2}\right)^{2}-\left(\sqrt{5}\right)^{2}$$

Expand \left(2r^{2}\right)^{2}.

$$2^{2}\left(r^{2}\right)^{2}-\left(\sqrt{5}\right)^{2}$$

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

$$2^{2}r^{4}-\left(\sqrt{5}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4r^{4}-\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$4r^{4}-5$$

Differentiate w.r.t. r

$$16r^{3}$$

Steps Using Derivative Rule for Sum

Consider \left(2r^{2}+\sqrt{5}\right)\left(2r^{2}-\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}r}(\left(2r^{2}\right)^{2}-\left(\sqrt{5}\right)^{2})$$

Expand \left(2r^{2}\right)^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}r}(2^{2}\left(r^{2}\right)^{2}-\left(\sqrt{5}\right)^{2})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}r}(2^{2}r^{4}-\left(\sqrt{5}\right)^{2})$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\mathrm{d}}{\mathrm{d}r}(4r^{4}-\left(\sqrt{5}\right)^{2})$$

The square of \sqrt{5} is 5.

$$\frac{\mathrm{d}}{\mathrm{d}r}(4r^{4}-5)$$

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}.

$$4\times 4r^{4-1}$$

Multiply 4 times 4.

$$16r^{4-1}$$

Subtract 1 from 4.

$$16r^{3}$$