Find the Derivative - d/dx f(x)=(2x-3)^4(x^2+x+1)^5

Problem

d()/d(x)*(2*x−3)^4*(x^2+x+1)^5

Solution

1. Identify the product rule for differentiation, which states that d()/d(x)*[u(x)*v(x)]=u(x)^′*v(x)+u(x)*v(x)^′

2. Assign the functions u(x)=(2*x−3)^4 and v(x)=(x^2+x+1)^5

3. Apply the chain rule to find u(x)^′=4*(2*x−3)^3⋅d(2*x−3)/d(x)=4*(2*x−3)^3*(2)=8*(2*x−3)^3

4. Apply the chain rule to find v(x)^′=5*(x^2+x+1)^4⋅d(x^2+x+1)/d(x)=5*(x^2+x+1)^4*(2*x+1)

5. Substitute these derivatives into the product rule formula.

ƒ(x)^′=8*(2*x−3)^3*(x^2+x+1)^5+(2*x−3)^4⋅5*(2*x+1)*(x^2+x+1)^4

6. Factor out the greatest common factor, which is (2*x−3)^3*(x^2+x+1)^4

ƒ(x)^′=(2*x−3)^3*(x^2+x+1)^4*[8*(x^2+x+1)+5*(2*x−3)*(2*x+1)]

7. Expand the terms inside the brackets.

ƒ(x)^′=(2*x−3)^3*(x^2+x+1)^4*[8*x^2+8*x+8+5*(4*x^2−4*x−3)]

8. Distribute the 5 and combine like terms.

ƒ(x)^′=(2*x−3)^3*(x^2+x+1)^4*[8*x^2+8*x+8+20*x^2−20*x−15]

ƒ(x)^′=(2*x−3)^3*(x^2+x+1)^4*(28*x^2−12*x−7)

Final Answer

d()/d(x)*(2*x−3)^4*(x^2+x+1)^5=(2*x−3)^3*(x^2+x+1)^4*(28*x^2−12*x−7)

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