differentiate\n⑤ $\frac{1}{sqrt{x^{3}+1}}$\n⑥ $x^{2}5^{-3x}$
differentiate\n⑤ $\frac{1}{sqrt{x^{3}+1}}$\n⑥ $x^{2}5^{-3x}$
Answer
Explanation:
Step1: Rewrite the function
Rewrite (\frac{1}{\sqrt{x^{3}+1}}) as ((x^{3}+1)^{-\frac{1}{2}}).
Step2: Apply the chain rule
The chain rule states that if (y = f(g(x))), then (y^\prime=f^\prime(g(x))\cdot g^\prime(x)). Let (u = x^{3}+1), so (y = u^{-\frac{1}{2}}). First, find (\frac{dy}{du}): [ \frac{dy}{du}=-\frac{1}{2}u^{-\frac{3}{2}} ] Then find (\frac{du}{dx}): [ \frac{du}{dx}=3x^{2} ] By the chain rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). Substitute (u = x^{3}+1) back in: [ \frac{dy}{dx}=-\frac{1}{2}(x^{3}+1)^{-\frac{3}{2}}\cdot3x^{2}=-\frac{3x^{2}}{2(x^{3}+1)^{\frac{3}{2}}} ]
Answer:
The derivative of (\frac{1}{\sqrt{x^{3}+1}}) is (-\frac{3x^{2}}{2(x^{3}+1)^{\frac{3}{2}}})