differentiate the function.\nf(x)=12(3x + 1)^{\\frac{2}{3}}(2x - 3)^{\\frac{5}{4}}\nf(x)=\\square
differentiate the function.\nf(x)=12(3x + 1)^{\\frac{2}{3}}(2x - 3)^{\\frac{5}{4}}\nf(x)=\\square
Answer
Explanation:
Step1: Apply the product rule
The product rule states that if (y = u\cdot v), then (y'=u'v + uv'). Let (u = 12(3x + 1)^{\frac{2}{3}}) and (v=(2x - 3)^{\frac{5}{4}}). First, find (u'): Using the chain rule ((f(g(x)))'=f'(g(x))\cdot g'(x)), where (f(t)=12t^{\frac{2}{3}}) and (t = 3x+1). (u'=12\times\frac{2}{3}(3x + 1)^{\frac{2}{3}-1}\times3=24(3x + 1)^{-\frac{1}{3}}) Second, find (v'): Using the chain rule, where (f(t)=t^{\frac{5}{4}}) and (t = 2x-3). (v'=\frac{5}{4}(2x - 3)^{\frac{5}{4}-1}\times2=\frac{5}{2}(2x - 3)^{\frac{1}{4}})
Step2: Calculate (f'(x))
By the product rule (f'(x)=u'v+uv') [ \begin{align*} f'(x)&=24(3x + 1)^{-\frac{1}{3}}(2x - 3)^{\frac{5}{4}}+12(3x + 1)^{\frac{2}{3}}\times\frac{5}{2}(2x - 3)^{\frac{1}{4}}\ &=24(3x + 1)^{-\frac{1}{3}}(2x - 3)^{\frac{5}{4}}+30(3x + 1)^{\frac{2}{3}}(2x - 3)^{\frac{1}{4}}\ &=6(3x + 1)^{-\frac{1}{3}}(2x - 3)^{\frac{1}{4}}\left[4(2x - 3)+5(3x + 1)\right]\ &=6(3x + 1)^{-\frac{1}{3}}(2x - 3)^{\frac{1}{4}}(8x-12 + 15x+5)\ &=6(3x + 1)^{-\frac{1}{3}}(2x - 3)^{\frac{1}{4}}(23x - 7) \end{align*} ]
Answer:
(f'(x)=6(3x + 1)^{-\frac{1}{3}}(2x - 3)^{\frac{1}{4}}(23x - 7))