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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Find the derivative of the function.

H ( r ) = ( r 2 1 ) 3 ( 2 r + 1 ) 5

To determine

To find: The derivative of H(r)=(r21)3(2r+1)5.

Explanation

Given:

The function is H(r)=(r21)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)]n1g(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((r21)3(2r+1)5)

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

H(r)=(2r+1)5ddr[(r21)3](r21)3ddr[(2r+1)5][(2r+1)5]2 (3)

Obtain the derivative ddr[(r21)3] by using the power rule combined with the chain rule as shown equation (1),

ddr[(r21)3]=3(r21)31ddr(r21)=3(r21)2(ddr(r2)ddr(1))=3(r21)2(2r0)=6r(r21)2

Thus, the derivative is ddr[(r21)3]=6r(r21)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)51ddr(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

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