Question

(f) $y = \frac{5e^{x^{2}}}{3x^{7}}$
(g) $y = e^{ln(x^{2}+3x)}$
(h) $h(x)=(9^{x})^{7}$
(i) $y = \log_{8}(2x + 1)$
(j) $g(x)=ln(\frac{5x^{2}}{e^{x}})$. note that there are multiple ways to take this involves using the chain rule then the quotient rule whereas another uses rules of logarithms to split apart the function first. try taking

1. determine where the tangent line to $f(x)=\frac{x}{(2x - 5)^{4}}$ is horizontal.
2. find the equation of the line tangent to $f(x)=sqrt{4x + 4}$ when $x = 3$.
3. expand $\log_{3}(5x^{2}y^{3}z^{4})$ as a sum of logarithms.
4. find $f(x)$ if $f(x)=7e^{x}-8ln(5x)-34$.

Explanation:

Step1: Identify problem type

These are calculus - related problems involving derivatives, tangent lines and logarithmic expansions.

Step2: Solve problem 11

1. Recall that the tangent line to $y = f(x)$ is horizontal when $f'(x)=0$.

• Use the quotient rule, if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = x$, $u'=1$, $v=(2x - 5)^{4}$, $v'=4(2x - 5)^{3}\times2=8(2x - 5)^{3}$.
• $f'(x)=\frac{1\times(2x - 5)^{4}-x\times8(2x - 5)^{3}}{(2x - 5)^{8}}=\frac{(2x - 5)^{3}[(2x - 5)-8x]}{(2x - 5)^{8}}=\frac{- 6x - 5}{(2x - 5)^{5}}$.
• Set $f'(x)=0$, then $-6x - 5 = 0$, so $x=-\frac{5}{6}$.

Step3: Solve problem 12

1. First, find the derivative of $y = f(x)=\sqrt{4x + 4}=(4x + 4)^{\frac{1}{2}}$.

• Using the chain - rule, if $y = u^{\frac{1}{2}}$ and $u = 4x+4$, then $y'=\frac{1}{2}(4x + 4)^{-\frac{1}{2}}\times4=\frac{2}{\sqrt{4x + 4}}$.
• When $x = 3$, $y=f(3)=\sqrt{4\times3 + 4}=\sqrt{16}=4$, and $y'|_{x = 3}=\frac{2}{\sqrt{4\times3+4}}=\frac{2}{4}=\frac{1}{2}$.
• Using the point - slope form of a line $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(3,4)$ and $m=\frac{1}{2}$, we get $y - 4=\frac{1}{2}(x - 3)$, or $y=\frac{1}{2}x+\frac{5}{2}$.

Step4: Solve problem 13

1. Use the product rule of logarithms $\log_{a}(MN)=\log_{a}M+\log_{a}N$ and $\log_{a}(M^{n})=n\log_{a}M$.

• $\log_{3}(5x^{2}y^{3}z^{4})=\log_{3}5+\log_{3}x^{2}+\log_{3}y^{3}+\log_{3}z^{4}=\log_{3}5 + 2\log_{3}x+3\log_{3}y + 4\log_{3}z$.

Step5: Solve problem 14

1. Use the derivative rules: $(e^{x})'=e^{x}$ and $(\ln u)'=\frac{u'}{u}$.

• If $f(x)=7e^{x}-8\ln(5x)-34$, then $f'(x)=7e^{x}-\frac{8\times5}{5x}=7e^{x}-\frac{8}{x}$.

Answer:

1. $x =-\frac{5}{6}$
2. $y=\frac{1}{2}x+\frac{5}{2}$
3. $\log_{3}5 + 2\log_{3}x+3\log_{3}y + 4\log_{3}z$
4. $f'(x)=7e^{x}-\frac{8}{x}$