Differentiate the function 𝑓(𝑥) = (3𝑥 + 1) 4 (2𝑥 − 1) 5 and simplify your answer
f(x)=(3x+1)4(2x−1)5f(x)=(3x+1)⁴(2x-1)⁵f(x)=(3x+1)4(2x−1)5
Let u=(3x+1)4=>dudx=12(3x+1)3u=(3x+1)⁴ =>\frac{du}{dx}=12(3x+1)³u=(3x+1)4=>dxdu=12(3x+1)3
v=(2x−1)5=>dvdx=10(2x−1)4v=(2x-1)⁵=>\frac{dv}{dx}=10(2x-1)⁴v=(2x−1)5=>dxdv=10(2x−1)4
f′(x)=vdudx+udvdxf'(x)=v\frac{du}{dx}+u\frac{dv}{dx}f′(x)=vdxdu+udxdv
f′(x)=(2x−1)5•12(3x+1)3+(3x+1)4•10(2x−1)4f'(x)=(2x-1)⁵•12(3x+1)³+(3x+1)⁴•10(2x-1)⁴f′(x)=(2x−1)5•12(3x+1)3+(3x+1)4•10(2x−1)4
f′(x)=2(2x−1)4(3x+1)3[6(2x−1)+5(3x+1)]f'(x)=2(2x-1)⁴(3x+1)³[6(2x-1)+5(3x+1)]f′(x)=2(2x−1)4(3x+1)3[6(2x−1)+5(3x+1)]
f′(x)=2(2x−1)4(3x+1)3[12x−6+15x+5]f'(x)=2(2x-1)⁴(3x+1)³[12x-6+15x+5]f′(x)=2(2x−1)4(3x+1)3[12x−6+15x+5]
f′(x)=2(2x−1)4(3x+1)3[27x−1]f'(x)=2(2x-1)⁴(3x+1)³[27x-1]f′(x)=2(2x−1)4(3x+1)3[27x−1]
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments