differentiate the function below.\n\n$y=(x^{2}+7)(x^{2}-3)^{13}$
differentiate the function below.\n\n$y=(x^{2}+7)(x^{2}-3)^{13}$
Answer
Explanation:
Step1: Apply the product rule
The product rule states that if (y = u\cdot v), then (y^\prime=u^\prime v + uv^\prime). Let (u=x^{2}+7) and (v=(x^{2}-3)^{13}). First, find (u^\prime): (u^\prime=\frac{d}{dx}(x^{2}+7)=2x).
Step2: Apply the chain rule to find (v^\prime)
The chain rule states that if (y = f(g(x))), then (y^\prime=f^\prime(g(x))\cdot g^\prime(x)). Let (g(x)=x^{2}-3) and (f(u)=u^{13}). Then (f^\prime(u) = 13u^{12}) and (g^\prime(x)=2x). So (v^\prime = 13(x^{2}-3)^{12}\cdot2x=26x(x^{2}-3)^{12}).
Step3: Substitute (u), (u^\prime), (v), and (v^\prime) into the product rule formula
(y^\prime=(2x)(x^{2}-3)^{13}+(x^{2}+7)\cdot26x(x^{2}-3)^{12}). Factor out (2x(x^{2}-3)^{12}): [ \begin{align*} y^\prime&=2x(x^{2}-3)^{12}[(x^{2}-3)+13(x^{2}+7)]\ &=2x(x^{2}-3)^{12}(x^{2}-3 + 13x^{2}+91)\ &=2x(x^{2}-3)^{12}(14x^{2}+88)\ &=4x(x^{2}-3)^{12}(7x^{2}+44) \end{align*} ]
Answer:
(y^\prime = 4x(x^{2}-3)^{12}(7x^{2}+44))