Question

27 - 76. calculate the derivative of the following functions.

1. $y=(3x^{2}+7x)^{10}$
2. $y=(x^{2}+2x + 7)^{8}$
3. $y=sqrt{10x + 1}$
4. $y=sqrt3{x^{2}+9}$

Explanation:

Step1: Use chain - rule for $y=(3x^{2}+7x)^{10}$

Let $u = 3x^{2}+7x$, then $y = u^{10}$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$. $\frac{dy}{du}=10u^{9}$ and $\frac{du}{dx}=6x + 7$. Then $\frac{dy}{dx}=10(3x^{2}+7x)^{9}(6x + 7)$.

Step2: Use chain - rule for $y=(x^{2}+2x + 7)^{8}$

Let $u=x^{2}+2x + 7$, then $y = u^{8}$. $\frac{dy}{du}=8u^{7}$ and $\frac{du}{dx}=2x + 2$. So $\frac{dy}{dx}=8(x^{2}+2x + 7)^{7}(2x + 2)=16(x + 1)(x^{2}+2x + 7)^{7}$.

Step3: Rewrite $y=\sqrt{10x + 1}=(10x + 1)^{\frac{1}{2}}$ and use chain - rule

Let $u = 10x+1$, then $y = u^{\frac{1}{2}}$. $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=10$. So $\frac{dy}{dx}=\frac{1}{2}(10x + 1)^{-\frac{1}{2}}\times10=\frac{5}{\sqrt{10x + 1}}$.

Step4: Rewrite $y=\sqrt[3]{x^{2}+9}=(x^{2}+9)^{\frac{1}{3}}$ and use chain - rule

Let $u=x^{2}+9$, then $y = u^{\frac{1}{3}}$. $\frac{dy}{du}=\frac{1}{3}u^{-\frac{2}{3}}$ and $\frac{du}{dx}=2x$. So $\frac{dy}{dx}=\frac{1}{3}(x^{2}+9)^{-\frac{2}{3}}\times2x=\frac{2x}{3\sqrt[3]{(x^{2}+9)^{2}}}$.

Answer:

1. $\frac{dy}{dx}=10(3x^{2}+7x)^{9}(6x + 7)$
2. $\frac{dy}{dx}=16(x + 1)(x^{2}+2x + 7)^{7}$
3. $\frac{dy}{dx}=\frac{5}{\sqrt{10x + 1}}$
4. $\frac{dy}{dx}=\frac{2x}{3\sqrt[3]{(x^{2}+9)^{2}}}$