2. use properties of logarithms and/or logarithmic differentiation to find the indicated derivatives a)…

2. use properties of logarithms and/or logarithmic differentiation to find the indicated derivatives a) $\frac{d}{dx}ln(\frac{3x}{sqrt{2x + 7}})$ b) $\frac{d}{dx}ln((x - 1)(5x + 2)(2x - 1))$ c) $y$ if $y = x^{cos x}$ d) $y$ if $y = x^{sqrt{x}}$

2. use properties of logarithms and/or logarithmic differentiation to find the indicated derivatives a) $\frac{d}{dx}ln(\frac{3x}{sqrt{2x + 7}})$ b) $\frac{d}{dx}ln((x - 1)(5x + 2)(2x - 1))$ c) $y$ if $y = x^{cos x}$ d) $y$ if $y = x^{sqrt{x}}$

Answer

Explanation:

Step1: Recall logarithmic properties

For part a, use $\ln(\frac{a}{b})=\ln a-\ln b$. So, $\ln(\frac{3x}{\sqrt{2x + 7}})=\ln(3x)-\ln(\sqrt{2x + 7})=\ln(3)+\ln(x)-\frac{1}{2}\ln(2x + 7)$. Differentiating term - by - term: The derivative of $\ln(3)$ is $0$ (constant), the derivative of $\ln(x)$ is $\frac{1}{x}$, and the derivative of $-\frac{1}{2}\ln(2x + 7)$ using the chain - rule is $-\frac{1}{2}\times\frac{2}{2x + 7}=-\frac{1}{2x + 7}$. So, $\frac{d}{dx}\ln(\frac{3x}{\sqrt{2x + 7}})=\frac{1}{x}-\frac{1}{2x + 7}=\frac{2x + 7-x}{x(2x + 7)}=\frac{x + 7}{x(2x + 7)}$.

Step2: For part b

Use $\ln(abc)=\ln a+\ln b+\ln c$. So, $\ln((x - 1)(5x + 2)(2x - 1))=\ln(x - 1)+\ln(5x + 2)+\ln(2x - 1)$. Differentiating term - by - term using the chain - rule: The derivative of $\ln(x - 1)$ is $\frac{1}{x - 1}$, the derivative of $\ln(5x + 2)$ is $\frac{5}{5x + 2}$, and the derivative of $\ln(2x - 1)$ is $\frac{2}{2x - 1}$. So, $\frac{d}{dx}\ln((x - 1)(5x + 2)(2x - 1))=\frac{1}{x - 1}+\frac{5}{5x + 2}+\frac{2}{2x - 1}$.

Step3: For part c

Take the natural logarithm of both sides: $\ln y=\cos x\ln x$. Differentiate both sides with respect to $x$ using the product - rule. The derivative of the left - hand side is $\frac{y'}{y}$ by the chain - rule. The derivative of the right - hand side: Using the product rule $(uv)' = u'v+uv'$ where $u=\cos x$ and $v = \ln x$. $u'=-\sin x$ and $v'=\frac{1}{x}$. So, $(\cos x\ln x)'=-\sin x\ln x+\frac{\cos x}{x}$. Then $y'=y(-\sin x\ln x+\frac{\cos x}{x})=x^{\cos x}(-\sin x\ln x+\frac{\cos x}{x})$.

Step4: For part d

Take the natural logarithm of both sides: $\ln y=\sqrt{x}\ln x$. Differentiate both sides with respect to $x$ using the product - rule. The derivative of the left - hand side is $\frac{y'}{y}$ by the chain - rule. For the right - hand side, using the product rule $(uv)' = u'v+uv'$ where $u = \sqrt{x}=x^{\frac{1}{2}}$ and $v=\ln x$. $u'=\frac{1}{2}x^{-\frac{1}{2}}$ and $v'=\frac{1}{x}$. $( \sqrt{x}\ln x)'=\frac{1}{2\sqrt{x}}\ln x+\frac{\sqrt{x}}{x}=\frac{\ln x}{2\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{\ln x + 2}{2\sqrt{x}}$. Then $y'=y\frac{\ln x + 2}{2\sqrt{x}}=x^{\sqrt{x}}\frac{\ln x+2}{2\sqrt{x}}$.

Answer:

a) $\frac{x + 7}{x(2x + 7)}$ b) $\frac{1}{x - 1}+\frac{5}{5x + 2}+\frac{2}{2x - 1}$ c) $x^{\cos x}(-\sin x\ln x+\frac{\cos x}{x})$ d) $x^{\sqrt{x}}\frac{\ln x + 2}{2\sqrt{x}}$