27 - 76. calculate the derivative of the following functions.\n27. $y=(3x^{2}+7x)^{10}$\n28. $y=(x^{2}+2x +…

Explanation:

Step1: Recall chain - rule

If $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$.

Step2: Solve for $y=(3x^{2}+7x)^{10}$

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

Step3: Solve for $y=\sqrt{10x + 1}=(10x + 1)^{\frac{1}{2}}$

Let $u = 10x+1$, so $y = u^{\frac{1}{2}}$. $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=10$. Then $\frac{dy}{dx}=\frac{1}{2}(10x + 1)^{-\frac{1}{2}}\cdot10=\frac{5}{\sqrt{10x + 1}}$.

Step4: Solve for $y = 5(7x^{3}+1)^{-3}$

Let $u = 7x^{3}+1$, so $y = 5u^{-3}$. $\frac{dy}{du}=-15u^{-4}$ and $\frac{du}{dx}=21x^{2}$. Then $\frac{dy}{dx}=-15(7x^{3}+1)^{-4}\cdot21x^{2}=-\frac{315x^{2}}{(7x^{3}+1)^{4}}$.

Step5: Solve for $y=\sec(3x + 1)$

Let $u = 3x + 1$, so $y=\sec u$. $\frac{dy}{du}=\sec u\tan u$ and $\frac{du}{dx}=3$. Then $\frac{dy}{dx}=3\sec(3x + 1)\tan(3x + 1)$.

Step6: Solve for $y=\tan(e^{x})$

Let $u = e^{x}$, so $y=\tan u$. $\frac{dy}{du}=\sec^{2}u$ and $\frac{du}{dx}=e^{x}$. Then $\frac{dy}{dx}=e^{x}\sec^{2}(e^{x})$.

Step7: Solve for $y=\sin(4x^{3}+3x + 1)$

Let $u = 4x^{3}+3x + 1$, so $y=\sin u$. $\frac{dy}{du}=\cos u$ and $\frac{du}{dx}=12x^{2}+3$. Then $\frac{dy}{dx}=(12x^{2}+3)\cos(4x^{3}+3x + 1)$.

Step8: Solve for $y=(5x + 1)^{\frac{2}{3}}$

Let $u = 5x + 1$, so $y = u^{\frac{2}{3}}$. $\frac{dy}{du}=\frac{2}{3}u^{-\frac{1}{3}}$ and $\frac{du}{dx}=5$. Then $\frac{dy}{dx}=\frac{10}{3(5x + 1)^{\frac{1}{3}}}$.

Step9: Solve for $y=\sqrt[4]{\frac{2x}{4x - 3}}=\left(\frac{2x}{4x - 3}\right)^{\frac{1}{4}}$

First, use the quotient - rule to find the derivative of $\frac{2x}{4x - 3}$. Let $f(x)=2x$ and $g(x)=4x - 3$, then $\left(\frac{f(x)}{g(x)}\right)^\prime=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{g(x)^{2}}=\frac{2(4x - 3)-2x\cdot4}{(4x - 3)^{2}}=\frac{8x-6 - 8x}{(4x - 3)^{2}}=-\frac{6}{(4x - 3)^{2}}$. Let $u=\frac{2x}{4x - 3}$, so $y = u^{\frac{1}{4}}$. $\frac{dy}{du}=\frac{1}{4}u^{-\frac{3}{4}}$. Then $\frac{dy}{dx}=\frac{1}{4}\left(\frac{2x}{4x - 3}\right)^{-\frac{3}{4}}\cdot\left(-\frac{6}{(4x - 3)^{2}}\right)=-\frac{3}{2(4x - 3)^{2}}\left(\frac{4x - 3}{2x}\right)^{\frac{3}{4}}$.

Step10: Solve for $y=(\sec x+\tan x)^{5}$

Let $u=\sec x+\tan x$, so $y = u^{5}$. $\frac{dy}{du}=5u^{4}$ and $\frac{du}{dx}=\sec x\tan x+\sec^{2}x$. Then $\frac{dy}{dx}=5(\sec x+\tan x)^{4}(\sec x\tan x+\sec^{2}x)$.

Step11: Solve for $y=(2x^{6}-3x^{3}+3)^{25}$

Let $u = 2x^{6}-3x^{3}+3$, so $y = u^{25}$. $\frac{dy}{du}=25u^{24}$ and $\frac{du}{dx}=12x^{5}-9x^{2}$. Then $\frac{dy}{dx}=25(2x^{6}-3x^{3}+3)^{24}(12x^{5}-9x^{2})$.

Answer:

For $y=(3x^{2}+7x)^{10}$, $y^\prime=10(3x^{2}+7x)^{9}(6x + 7)$; For $y=\sqrt{10x + 1}$, $y^\prime=\frac{5}{\sqrt{10x + 1}}$; For $y = 5(7x^{3}+1)^{-3}$, $y^\prime=-\frac{315x^{2}}{(7x^{3}+1)^{4}}$; For $y=\sec(3x + 1)$, $y^\prime=3\sec(3x + 1)\tan(3x + 1)$; For $y=\tan(e^{x})$, $y^\prime=e^{x}\sec^{2}(e^{x})$; For $y=\sin(4x^{3}+3x + 1)$, $y^\prime=(12x^{2}+3)\cos(4x^{3}+3x + 1)$; For $y=(5x + 1)^{\frac{2}{3}}$, $y^\prime=\frac{10}{3(5x + 1)^{\frac{1}{3}}}$; For $y=\sqrt[4]{\frac{2x}{4x - 3}}$, $y^\prime=-\frac{3}{2(4x - 3)^{2}}\left(\frac{4x - 3}{2x}\right)^{\frac{3}{4}}$; For $y=(\sec x+\tan x)^{5}$, $y^\prime=5(\sec x+\tan x)^{4}(\sec x\tan x+\sec^{2}x)$; For $y=(2x^{6}-3x^{3}+3)^{25}$, $y^\prime=25(2x^{6}-3x^{3}+3)^{24}(12x^{5}-9x^{2})$; (Answers for other functions can be derived in a similar chain - rule fashion)