To determine

To find:  The derivative of f(x)=(2x−3)4(x2+x+1)5.

Answer

The derivative of f(x)=(2x−3)4(x2+x+1)5 is (2x−3)3(x2+x+1)4(28x2−12x−7)_.

Explanation

Given:

The function is f(x)=(2x−3)4(x2+x+1)5.

Result used:

The Power Rule combined with the Chain Rule:

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

ddx[g(x)]n=n[g(x)]n−1⋅g′(x) (1)

Product Rule:

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

ddx[f(x)g(x)]=f(x)ddx[g(x)]+g(x)ddx[f(x)] (2)

Calculation:

Obtain the derivative of f(x).

f′(x)=ddx(f(x))=ddx((2x−3)4(x2+x+1)5)

Apply the product rule as shown in equation (2),

f′(x)=(2x−3)4⋅ddx((x2+x+1)5)+(x2+x+1)5⋅ddx((2x−3)4) (3)

Obtain the derivative ddx((x2+x+1)5) by using the power rule combined with the chain rule as shown equation (1).

ddx((x2+x+1)5)=5(x2+x+1)5−1⋅ddx(x2+x+1)=5(x2+x+1)4(ddx(x2)+ddx(x)+ddx(1))=5(x2+x+1)4((2x2−1)+(1x1−1)+(0))=5(x2+x+1)4((2x)+(1))

                           =5(x2+x+1)4(2x+1)

Thus, the derivative is ddx((x2+x+1)5)=5(x2+x+1)4((2x)+(1)) (4)

Obtain the derivative ddx(2x−3)4 by using the power rule combined with the chain rule as shown equation (1).

ddx(2x−3)4=4(2x−3)4−1⋅ddx(2x−3)=4(2x−3)3(ddx(2x)−ddx(3))=4(2x−3)3(2ddx(x)−ddx(3))=4(2x−3)3(2(1x1−1)−0)

Simplify further, the derivative becomes

ddx(2x−3)4=4(2x−3)3(2(1))=4(2x−3)3(2)=8(2x−3)3

Thus, the derivative is ddx(2x−3)4=8(2x−3)3 (5)

Substitute equations (4) and (5) in equation (3),

f′(x)=(2x−3)4[5(x2+x+1)4(2x+1)]+(x2+x+1)5[8(2x−3)3]=5(2x−3)4(x2+x+1)4(2x+1)+8(x2+x+1)5(2x−3)3=(2x−3)3(x2+x+1)4(5(2x−3)(2x+1)+8(x2+x+1))=(2x−3)3(x2+x+1)4((10x−15)(2x+1)+(8x2+8x+8))

Simplify further and obtain the derivative of the function.

f′(x)=(2x−3)3(x2+x+1)4(20x2+10x−30x−15+8x2+8x+8)=(2x−3)3(x2+x+1)4(28x2−12x−7)

Therefore, the derivative of f(x)=(2x−3)4(x2+x+1)5 is f′(x)=(2x−3)3(x2+x+1)4(28x2−12x−7)_.