#### To determine

**To find:** The derivative of H(r)=(r2−1)3(2r+1)5.

#### Answer

The derivative of H(r)=(r2−1)3(2r+1)5 is H′(r)=2(r2−1)2[r2+3r+5](2r+1)6_.

#### Explanation

**Given:**

The function is H(r)=(r2−1)3(2r+1)5.

**Result used: The Power Rule combined with the Chain Rule**

If *n* is any real number and g(r) is differentiable function, then

ddr[g(r)]n=n[g(r)]n−1g′(r) (1)

**Quotient Rule:**

If f(r) and g(r) are both differentiable function, then

ddr[f(r)g(r)]=g(r)ddr[f(r)]−f(r)ddr[g(r)][g(r)]2 (2)

**Calculation:**

Obtain the derivative of H(r).

H′(r)=ddx(H(r))=ddx((r2−1)3(2r+1)5)

Apply the Quotient Rule as shown in equation (2),

H′(r)=(2r+1)5ddr[(r2−1)3]−(r2−1)3ddr[(2r+1)5][(2r+1)5]2 (3)

Obtain the derivative ddr[(r2−1)3] by using the power rule combined with the chain rule as shown equation (1),

ddr[(r2−1)3]=3(r2−1)3−1ddr(r2−1)=3(r2−1)2(ddr(r2)−ddr(1))=3(r2−1)2(2r−0)=6r(r2−1)2

Thus, the derivative is ddr[(r2−1)3]=6r(r2−1)2.

Obtain the derivative ddr[(2r+1)5] by using the power rule combined with the chain rule as shown equation (1),

ddr[(2r+1)5]=5(2r+1)5−1ddr(2r+1)=5(2r+1)4(ddr(2r)+ddr(1))=5(2r+1)4(2+0)=10(2r+1)4

Thus, the derivative is ddr[(2r+1)5]=10(2r+1)4.

Substitute 6r(r2−1)2 for ddr[(r2−1)3] and 10(2r+1)4 for ddr[(2r+1)5] in

equation (3),

H′(r)=(2r+1)5(6r(r2−1)2)−(r2−1)3(10(2r+1)4)[(2r+1)5]2=(2r+1)4(r2−1)2[(6r(2r+1))−(r2−1)(10)](2r+1)10=(r2−1)2[((12r2+6r))−(10r2−10)](2r+1)−4(2r+1)10=(r2−1)2[12r2+6r−10r2+10](2r+1)10−4

Simplify further and obtain the derivative.

H′(r)=(r2−1)2[2r2+6r+10](2r+1)6=2(r2−1)2[r2+3r+5](2r+1)6

Therefore, the derivative of H(r)=(r2−1)3(2r+1)5 is H′(r)=2(r2−1)2[r2+3r+5](2r+1)6_.