Question

Differentiate the following expressions
(i) \[\log \left[ {{a}^{4x}}{{\left( \dfrac{x-5}{x+4} \right)}^{\dfrac{3}{4}}} \right]\]
(ii) \[\log [{{\sin }^{3}}x.{{\cos }^{4}}x.{{({{x}^{2}}-1)}^{5}}]\]
(iii) \[\log \left[ \dfrac{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}{-x+\sqrt{{{x}^{2}}+{{a}^{2}}}} \right]\]
(iv) \[{{4}^{{{\log }_{2}}\sec x}}-{{9}^{{{\log }_{3}}\tan x}}\]

Answer 1

Hint: We will first use logarithmic rules to first simplify the expressions and then we will differentiate to get the answers. So of the logarithmic rules are \[\log (AB)=\log A+\log B\], \[\log {{x}^{n}}=n\log x\] and \[\log \left( \dfrac{A}{B} \right)=\log A-\log B\].

Complete step-by-step solution -
(i) \[y=\log \left[ {{a}^{4x}}{{\left( \dfrac{x-5}{x+4} \right)}^{\dfrac{3}{4}}} \right].........(1)\]
First we will simplify equation (1) by applying the product rule that is \[\log (AB)=\log A+\log B\].
\[y=\log {{a}^{4x}}+\log {{\left( \dfrac{x-5}{x+4} \right)}^{\dfrac{3}{4}}}..........(2)\]
Now applying \[\log {{x}^{n}}=n\log x\] rule in equation (2) we get,
\[y=4x\log a+\dfrac{3}{4}\log \left( \dfrac{x-5}{x+4} \right)..........(3)\]
Now differentiating both sides of equation (3) we get,
\[\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ 4x\log a+\dfrac{3}{4}\log \left( \dfrac{x-5}{x+4} \right) \right]..........(4)\]
We know that differentiation of a constant is zero and also the differentiation of log(x) is \[\dfrac{1}{x}\]. We also know \[\log \left( \dfrac{A}{B} \right)=\log A-\log B\]. So using these information in equation (4) we get,
\[\begin{align}
  & \dfrac{dy}{dx}=\left[ 4\log a\dfrac{d}{dx}(x)+\dfrac{3}{4}\dfrac{d}{dx}[\log \left( x-5 \right)]-\dfrac{3}{4}\dfrac{d}{dx}[\log \left( x+4 \right)] \right] \\
 & \dfrac{dy}{dx}=\left[ 4\log a.1+\dfrac{3}{4}\times \dfrac{1}{x-5}\dfrac{d}{dx}(x-5)-\dfrac{3}{4}\times \dfrac{1}{x+4}\dfrac{d}{dx}(x+4) \right].......(5) \\
\end{align}\]
Now differentiating again in equation (5) we get the final answer,
\[\begin{align}
  & \dfrac{dy}{dx}=\left[ 4\log a+\dfrac{3}{4}\times \dfrac{1}{x-5}(1-0)-\dfrac{3}{4}\times \dfrac{1}{x+4}(1+0) \right] \\
 & \dfrac{dy}{dx}=4\log a+\dfrac{3}{4(x-5)}-\dfrac{3}{4(x+4)} \\
\end{align}\]
Hence the answer is \[\dfrac{dy}{dx}=4\log a+\dfrac{3}{4(x-5)}-\dfrac{3}{4(x+4)}\].

(ii) \[y=\log [{{\sin }^{3}}x.{{\cos }^{4}}x.{{({{x}^{2}}-1)}^{5}}].......(1)\]
First we will simplify equation (1) by applying the product rule that is \[\log (ABC)=\log A+\log B+\log C\].
\[y=\log {{\sin }^{3}}x+\log {{\cos }^{4}}x+\log {{({{x}^{2}}-1)}^{5}}.......(2)\]
Now applying \[\log {{x}^{n}}=n\log x\] rule in equation (2) we get,
\[y=3\log \sin x+4\log \cos x+5\log ({{x}^{2}}-1).......(3)\]
Now differentiating both sides of equation (3) we get,
\[\dfrac{dy}{dx}=3\dfrac{d}{dx}(\log \sin x)+4\dfrac{d}{dx}(\log \cos x)+5\dfrac{d}{dx}[\log ({{x}^{2}}-1)].......(4)\]
We know that differentiation of a constant is zero and also differentiation of log(x) is \[\dfrac{1}{x}\]. So using this information in equation (4) we get,
\[\dfrac{dy}{dx}=3\times \dfrac{1}{\sin x}\dfrac{d}{dx}(\sin x)+4\times \dfrac{1}{\cos x}\dfrac{d}{dx}(\cos x)+5\times \dfrac{1}{{{x}^{2}}-1}\dfrac{d}{dx}[({{x}^{2}}-1)].......(5)\]
Now differentiating again in equation (5) we get the final answer,
\[\begin{align}
  & \dfrac{dy}{dx}=3\times \dfrac{1}{\sin x}\times \cos x+4\times \dfrac{1}{\cos x}\times -\sin x+5\times \dfrac{1}{{{x}^{2}}-1}\times 2x \\
 & \dfrac{dy}{dx}=3\cot x-4\tan x+\dfrac{10x}{{{x}^{2}}-1} \\
\end{align}\]
Hence the answer is \[\dfrac{dy}{dx}=3\cot x-4\tan x+\dfrac{10x}{{{x}^{2}}-1}\].

(iii) \[y=\log \left[ \dfrac{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}{-x+\sqrt{{{x}^{2}}+{{a}^{2}}}} \right]...........(1)\]
First we will simplify equation (1) by applying the division rule that is \[\log \left( \dfrac{A}{B} \right)=\log A-\log B\].
\[y=\log (x+\sqrt{{{x}^{2}}+{{a}^{2}}})-\log (-x+\sqrt{{{x}^{2}}+{{a}^{2}}})...........(2)\]
Now differentiating both sides of equation (2) we get,
\[\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ \log (x+\sqrt{{{x}^{2}}+{{a}^{2}}}) \right]-\dfrac{d}{dx}\left[ \log (-x+\sqrt{{{x}^{2}}+{{a}^{2}}}) \right]...........(3)\]
We know that differentiation of a constant is zero and also differentiation of log(x) is \[\dfrac{1}{x}\]. So using this information in equation (3) we get,
\[\dfrac{dy}{dx}=\dfrac{1}{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}\times \dfrac{d}{dx}\left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]-\dfrac{1}{-x+\sqrt{{{x}^{2}}+{{a}^{2}}}}\dfrac{d}{dx}\left[ -x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]...........(4)\]
Now differentiating again in equation (4) we get,
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{1}{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}\times \left[ 1+\dfrac{1}{2\sqrt{{{x}^{2}}+{{a}^{2}}}}\dfrac{d}{dx}({{x}^{2}}+{{a}^{2}}) \right]-\dfrac{1}{-x+\sqrt{{{x}^{2}}+{{a}^{2}}}}\times \left[ -1+\dfrac{1}{2\sqrt{{{x}^{2}}+{{a}^{2}}}}\dfrac{d}{dx}({{x}^{2}}+{{a}^{2}}) \right] \\
 & \dfrac{dy}{dx}=\dfrac{1}{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}\times \left[ 1+\dfrac{2x}{2\sqrt{{{x}^{2}}+{{a}^{2}}}} \right]-\dfrac{1}{-x+\sqrt{{{x}^{2}}+{{a}^{2}}}}\times \left[ -1+\dfrac{2x}{2\sqrt{{{x}^{2}}+{{a}^{2}}}} \right] \\
 & \dfrac{dy}{dx}=\dfrac{1}{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}\times \left[ \dfrac{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right]-\dfrac{1}{-(x-\sqrt{{{x}^{2}}+{{a}^{2}}})}\times \left[ \dfrac{x-\sqrt{{{x}^{2}}+{{a}^{2}}}}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right].........(5) \\
 & \\
\end{align}\]
Cancelling similar terms from equation (5) we get the final answer,
\[\dfrac{dy}{dx}=\dfrac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}+\dfrac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}=\dfrac{2}{\sqrt{{{x}^{2}}+{{a}^{2}}}}\]
Hence the answer is \[\dfrac{dy}{dx}=\dfrac{2}{\sqrt{{{x}^{2}}+{{a}^{2}}}}\].

(iv) \[y={{4}^{{{\log }_{2}}\sec x}}-{{9}^{{{\log }_{3}}\tan x}}........(1)\]
4 can be written as \[{{2}^{2}}\] and 9 can be written as \[{{3}^{2}}\]. So using this in equation (1) we get,
\[y={{2}^{2{{\log }_{2}}\sec x}}-{{3}^{3{{\log }_{3}}\tan x}}........(2)\]
Now applying \[\log {{x}^{n}}=n\log x\] rule in equation (2) we get,
\[y={{2}^{{{\log }_{2}}{{\sec }^{2}}x}}-{{3}^{{{\log }_{3}}{{\tan }^{2}}x}}........(3)\]
We know that \[{{a}^{{{\log }_{a}}x}}=x\]. Now applying this rule in equation (3) we get,
\[\begin{align}
  & y={{\sec }^{2}}x-{{\tan }^{2}}x \\
 & y=1......(4) \\
\end{align}\]
Now differentiating equation (4) we get,
\[\dfrac{dy}{dx}=0\].
Hence the answer is \[\dfrac{dy}{dx}=0\].

Note: Remembering logarithmic rules and trigonometric identities is the key here. Also we should know the different differentiation formulas. And differentiation of a constant is always zero.