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Derivatives Of Log Functions

Learn derivatives of logarithmic functions and logarithmic differentiation examples, finding the derivative of expressions and equation of tangent to a curve involving log function at particular values of x.

Derivative of Logarithmic Functions

  • If \(y=\ell n\,x\), then \(\dfrac {dy}{dx}=\dfrac {1}{x}\) 

           \(\Rightarrow \dfrac {d}{dx}\,(\ell n\,x)=\dfrac {1}{x}\)

  • The derivative of \(\ell n\,x\) w.r.t. \(x\) is  \(\dfrac {1}{x}\).
  • If \(u(x)\) is a function of x, then \(\dfrac {d}{dx}\,(\ell n\,u)=\dfrac {1}{u}\,\dfrac {du}{dx}\)    (by Chain Rule)

e.g. \(\dfrac {d}{dx}\Big(\ell n\,(cot\,x)\Big)=\dfrac {1}{cot\,x}×\dfrac {d}{dx}\,cot\,x\)

\(=\dfrac {1}{cot\,x}×(-cosec^2x)=\dfrac {-1}{sin\,x\,cosx}\)

  • The above can also be written as :\(\dfrac {d}{dx}\left (\ell n (g(x)\right)=\dfrac {1}{g(x)}×g'(x)\)

 

Illustration Questions

If  \(y=\ell n\,(2x^2+3x+1)\), then \(\dfrac {dy}{dx}\) is the expression

A \(\dfrac {8x+5}{2x^2+3x+1}\)

B \(\dfrac {2x+7}{x^2+x+1}\)

C \(\ell n\,x+e^x\)

D \(\dfrac {4x+3}{2x^2+3x+1}\)

×

\(y=\ell n (2x^2+3x+1)=\ell n \,u \;\;\;\;\;\;\;\;\;\;\; (Let\;\;u=2x^2+3x+1)\)

\(\dfrac {d}{dx}\left (\ell n\,u\right)=\dfrac {1}{u}\;\dfrac {du}{dx}\)

\(\therefore \dfrac {dy}{dx}=\dfrac {1}{u}\;\dfrac {du}{dx}=\dfrac {1}{2x^2+3x+1}×\dfrac {d}{dx}(2x^2+3x+1)\)

\(=\dfrac {4x+3}{2x^2+3x+1}\)

If  \(y=\ell n\,(2x^2+3x+1)\), then \(\dfrac {dy}{dx}\) is the expression

A

\(\dfrac {8x+5}{2x^2+3x+1}\)

.

B

\(\dfrac {2x+7}{x^2+x+1}\)

C

\(\ell n\,x+e^x\)

D

\(\dfrac {4x+3}{2x^2+3x+1}\)

Option D is Correct

Finding the Derivative of Expressions Involving Log Function at Particular Values of x

  • \(f'(a)\) means the value of derivative of the function \('f'\) at \(x=a\).

Illustration Questions

If  \(f(x)=\ell n\,\left (\sqrt x +\dfrac {1}{\sqrt x}\right)\), then the value of \(f'(3)\) is

A \(\dfrac {1}{12}\)

B \(\dfrac {1}{4}\)

C \(\dfrac {5}{6}\)

D \(\dfrac {7}{12}\)

×

\(f(x)=\ell n\,\left (\sqrt x +\dfrac {1}{\sqrt x}\right)\)\(=\ell n\,u\;\;\;\;\;\;\;\left (Let\;\;\;\;\;\;u=\sqrt x +\dfrac {1}{\sqrt x}\right)\)

\(\therefore \; f'(x)=\dfrac {1}{\sqrt x +\dfrac {1}{\sqrt x}}× \left (\dfrac {d}{x} \left ( \sqrt x +\dfrac {1}{\sqrt x}\right) \right) \)

\(= \dfrac {\sqrt x}{x+1}× \left ( \dfrac {1}{2\sqrt x}+\dfrac {-1}{2}x^{-3/2} \right)\)

\(= \dfrac {\sqrt x}{x+1}× \left ( \dfrac {1}{2\sqrt x}-\dfrac {1}{2x\sqrt x} \right)\)

\(\therefore f'(3)= \dfrac {\sqrt 3}{4}× \left ( \dfrac {1}{2\sqrt 3}-\dfrac {1}{6\sqrt 3} \right)\)

\(f'(3)= \dfrac {\sqrt 3}{4}× \dfrac {2}{6\sqrt 3}=\dfrac {1}{12}\)

If  \(f(x)=\ell n\,\left (\sqrt x +\dfrac {1}{\sqrt x}\right)\), then the value of \(f'(3)\) is

A

\(\dfrac {1}{12}\)

.

B

\(\dfrac {1}{4}\)

C

\(\dfrac {5}{6}\)

D

\(\dfrac {7}{12}\)

Option A is Correct

Finding the Equation of Tangent to a Curve which contain Log Function

  • Equation of tangent to \(y=f(x)\) at a point  \((a, f(a))\) on it is \(y-f(a)=f'(a)(x-a)\)

Illustration Questions

Find the equation of tangent to the curve  \(y=(\ell n\,x)×x^3\) at the point (1, 0) on it.

A \(x+2y-1=0\)

B \(5x-3y-7=0\)

C \(x-y-1=0\)

D \(9x+y-9=0\)

×

\(y=x^3\;\ell n\,x\)

\(\Rightarrow \dfrac {dy}{dx}=f'(x)=x^3×\dfrac {1}{x}+3x^2\,\ell n\,x\)

\(=x^2+3x^2\;\ell n\,x\)

Equation of tangent at (1, 0) is

\(y-0=f'(1)(x-1)\)

\(\Rightarrow y=f'(1)(x-1)\)

\(f'(1)=1+3×1×\ell n\,1=1+0=1\)

\(\therefore\) Equation of tangent is

\(y=1(x-1)\)

\(\Rightarrow x-y-1=0\)

Find the equation of tangent to the curve  \(y=(\ell n\,x)×x^3\) at the point (1, 0) on it.

A

\(x+2y-1=0\)

.

B

\(5x-3y-7=0\)

C

\(x-y-1=0\)

D

\(9x+y-9=0\)

Option C is Correct

Derivative of Logarithmic Functions where Base is any allowed Quantity 'a'

  • \(\dfrac {d}{dx}(log_a\,x)=\dfrac {d}{dx}\underbrace {\left (\dfrac {\ell n\, x}{\ell n\,a}\right)}_{\text {Base change formula}} =\dfrac {1}{\ell n\,a}×\dfrac {d}{dx}(\ell n\,x)=\dfrac {1}{x\,\ell n\,a}\)

\(\therefore \;\dfrac {d}{dx}(log_a\,x)=\dfrac {1}{x\,\ell n\,a}\)

  • If we take \(a=e\), then it reduces to \(\dfrac {d}{dx}(\ell n\,x)=\dfrac {1}{x}\)

\( \;\dfrac {d}{dx}(log_a\,f(x))=\underbrace{\dfrac {1}{\ell n\,a}× \dfrac {1}{f(x)}×f'(x)=\dfrac {f'(x)}{f(x)\,\ell n\,a}}_{chain\, rule} \)

\(\therefore \;\dfrac {d}{dx}(log_a\,f(x)) ={\dfrac {1}{f(x)\,\ell n\,a}× f'(x)}\)

Illustration Questions

If \(y=log_2\,(e^{2x}\,sin\,x)\), then find \( \;\dfrac {dy}{dx}\).

A \(\ell n\,x-x^2+x^3+C\)

B \(\dfrac {(cos\,x+2sin\,x)}{(\ell n\,2)(sin\,x)}\)

C \(\dfrac {(2\,cos\,x-sin\,x)}{(\ell n\,2)(cos\,x)}\)

D \(e^{log^{(sin\,x)}}+cos\,x+C\)

×

\(y=log_2(e^{2x}\,sin\,x)\)

\( \;\dfrac {dy}{dx}=\dfrac {1}{\ell n\,2}× \dfrac {1}{e^{2x}\,sin\,x} ×\dfrac {d}{dx}(e^{2x}\,sin\,x)\)

\(\therefore \;\dfrac {dy}{dx}=\dfrac {1}{(\ell n\,2) ({e^{2x}}\,sin\,x)} ×(e^{2x}\,cos\,x+2e^{2x}\,sin\,x)\)

\(=\dfrac {e^{2x}(cos\,x+2sin\,x)}{(\ell n\,2) ({e^{2x}}\,sin\,x)} =\dfrac {cosx+2sin\,x}{\ell n\,2(sin\,x)}\)

If \(y=log_2\,(e^{2x}\,sin\,x)\), then find \( \;\dfrac {dy}{dx}\).

A

\(\ell n\,x-x^2+x^3+C\)

.

B

\(\dfrac {(cos\,x+2sin\,x)}{(\ell n\,2)(sin\,x)}\)

C

\(\dfrac {(2\,cos\,x-sin\,x)}{(\ell n\,2)(cos\,x)}\)

D

\(e^{log^{(sin\,x)}}+cos\,x+C\)

Option B is Correct

Logarithmic Differentiation

  • Sometimes we face complicated functions which involve products, quotients or powers of some terms, these can often be simplified by taking logarithm and then differentiated. This is called logarithmic differentiation.
  • e.g.

               If  \(y=\dfrac {(x+1)^3×\sqrt [4] {x-2}}{\sqrt [5] {(x-3)^2}} \) and we have to find \(\dfrac{dy}{dx}\).

We take log both sides to base e.

\(\therefore\ell n\;y=\ell n\left(\dfrac {(x+1)^3×\sqrt [4] {x-2}}{\sqrt [5] {(x-3)^2}}\right) \)  (Now, use properties of log)

\(=3\,\ell n (x+1)+\dfrac {1}{4}\ell n(x-2)-\dfrac {2}{5}\ell n\, (x-3)\)

Now, observe that we can differentiate both sides easily.

\(\Rightarrow \dfrac {1}{y}\,\dfrac {dy}{dx}=\dfrac {3}{x+1}+\dfrac {1}{4(x-2)}-\dfrac {2}{5(x-3)}\)

\(\Rightarrow \dfrac {dy}{dx}=y\left [\dfrac {3}{x+1}+\dfrac {1}{4(x-2)}-\dfrac {2}{5(x-3)}\right]\)

\(=\dfrac {(x+1)^3×\sqrt [4] {x-2}}{\sqrt [5] {(x-3)^2}} \left [\dfrac {3}{x+1}+\dfrac {1}{4(x-2)}-\dfrac {2}{5(x-3)}\right]\)

 

Illustration Questions

If  \(y=\dfrac {(x-2)^2×\sqrt [3] {(x+1)}}{(x-5)^3} \), then find \(\dfrac {dy}{dx}\).

A \(\dfrac {(x-2)^2×(x+1)^{1/3}}{(x-5)^3} \left [\dfrac {2}{x-2}+\dfrac {1}{3(x+1)}-\dfrac {3}{x-5}\right]\)

B \(3\,sin\,x-2cos^2\,x+\dfrac {1}{x}\)

C \(\dfrac {(x-2)^2×\sqrt [3] {x+1}}{ {(x-5)^3}} \left [\dfrac {3}{x-2}+\dfrac {1}{x-5}\right]\)

D \(\dfrac {3}{sin\,x}-\dfrac {2}{x+1}+cos^3\,x\)

×

\(y=\dfrac {(x-2)^2×\sqrt [3] {(x+1)}}{(x-5)^3} \)

Take log both sides to the base e.

\(\Rightarrow \;\ell n\,y=\ell n \left (\dfrac {(x-2)^2×(x+1)^{1/3}}{(x-5)^3}\right) \)

\(\Rightarrow \;\ell n\,(x-2)^2+\ell n(x+1)^{1/3}-\ell n(x-5)^{3}\)

\(\Rightarrow \;2\,\ell n\,(x-2)+\dfrac {1}{3}\ell n(x+1)-3\,\ell n(x-5)\)

Now, differentiate both sides with respect to x

\(\Rightarrow \dfrac {1}{y}\,\dfrac {dy}{dx}=\dfrac {2}{x-2}+\dfrac {1}{3(x+1)}-\dfrac {3}{x-5}\)

\(\Rightarrow \dfrac {1}{y}\,\dfrac {dy}{dx}=\dfrac {(x-2)^2\,(x+1)^{1/3}}{(x-5)^3} \left [\dfrac {2}{x-2}+\dfrac {1}{3(x+1)}-\dfrac {3}{x-5}\right]\)

If  \(y=\dfrac {(x-2)^2×\sqrt [3] {(x+1)}}{(x-5)^3} \), then find \(\dfrac {dy}{dx}\).

A

\(\dfrac {(x-2)^2×(x+1)^{1/3}}{(x-5)^3} \left [\dfrac {2}{x-2}+\dfrac {1}{3(x+1)}-\dfrac {3}{x-5}\right]\)

.

B

\(3\,sin\,x-2cos^2\,x+\dfrac {1}{x}\)

C

\(\dfrac {(x-2)^2×\sqrt [3] {x+1}}{ {(x-5)^3}} \left [\dfrac {3}{x-2}+\dfrac {1}{x-5}\right]\)

D

\(\dfrac {3}{sin\,x}-\dfrac {2}{x+1}+cos^3\,x\)

Option A is Correct

Solving Function of the form f(x)g(x) by Logarithmic Differentiation

  • Sometimes we come across functions of the form  \(y=(f(x))^{g(x)}\), to differentiate function of these form we first take log on both sides and then differentiate.
  • \(y=(f(x))^{g(x)}\Rightarrow\) Take log on both sides to base e

\(\Rightarrow \ell n\,y=\ell n(f(x))^{g(x)}\)

\(\Rightarrow \ell n\,y=g(x)\;\ell n(f(x))\)

\(\Rightarrow \dfrac {1}{y}\,\dfrac {dy}{dx}=\underbrace{g'(x)\,\ell n\,f(x)+\dfrac {g(x)}{f(x)}f'(x)}_{Product\, rule}\)

\(\Rightarrow \dfrac {dy}{dx}=(f(x))^{g(x)}\Big[g'(x)\,\ell n\,f(x)+\dfrac {g(x)}{f(x)}f'(x)\Big]\)?

  • Do not remember the formula but do these steps for the \('f'\)  and \('g'\) given in the problem.

 

Illustration Questions

If \(y=(cos\,x)^{x^2}\), find \(\dfrac {dy}{dx}\).

A \((cos\,x)^{x^2} [-x^2\,tan\,x+2x\,ln\,(cos\,x)]\)

B \((cos\,x)^{x^2} [-x\,tan\,x+x\,\ell n(cosx)]\)

C \((sin\,x)^{cos\,x}\,(1+2x)\)

D \((tan\,x)\,(e^x+\ell n\,x)\)

×

\(y=(cos\,x)^{x^2}\rightarrow\) of the form \((f(x))^{g(x)}\)

Take log on both sides to the base e

\(\Rightarrow \ell n\,y=\ell n(cos\,x)^{x^2}\)

\(\Rightarrow \ell n\,y=x^2\,\ell n (cos\,x)\)

Now differentiate both sides with respect to x ,

\(\Rightarrow \dfrac {1}{y}\,\dfrac {dy}{dx}=x^2\dfrac {d}{dx}(\ell n\,(cos\,x))+ \,\ell n\,(cos\,x) \dfrac {d}{dx}\,x^2\)

\(\Rightarrow \dfrac {1}{y}\,\dfrac {dy}{dx}=x^2×\dfrac {1}{cos\,x}×(-sin\,x)+2x\;\ell n\,(cos\,x)\)

\(\Rightarrow \dfrac {dy}{dx}=(cos\,x)^{x^2}[-x^2\,tan\,x+2x\,\ell n\,(cos\,x)]\)

If \(y=(cos\,x)^{x^2}\), find \(\dfrac {dy}{dx}\).

A

\((cos\,x)^{x^2} [-x^2\,tan\,x+2x\,ln\,(cos\,x)]\)

.

B

\((cos\,x)^{x^2} [-x\,tan\,x+x\,\ell n(cosx)]\)

C

\((sin\,x)^{cos\,x}\,(1+2x)\)

D

\((tan\,x)\,(e^x+\ell n\,x)\)

Option A is Correct

Higher Order Derivatives of Functions which Involve Logarithmic Expression

  • \(f''(x)=\dfrac {d^2y}{dx^2}=\dfrac {d}{dx}\left (\dfrac {dy}{dx}\right)\)
  • In general, \(f^n(x)=\dfrac {d^ny}{dx^n}=n^{th}\) derivative of \(f\) with respect to \(x\).
  • Treat \(f'\) as a function , differentiate it again to get \(f''(x)\) and so on.

 

 

Illustration Questions

If \(y=x^3\,\ell n\,x\), find \(\dfrac {d^2y}{dx^2}\).

A \(x(5+6\,\ell n\,x)\)

B \(x^2(6+5\,\ell n\,x)\)

C \((sin\,x)(2+x^2)\)

D \(\dfrac {\ell n\,x}{x}\)

×

\(y=x^3\,\ell n\,x\)

\(\dfrac {dy}{dx}=x^3\,×\dfrac {1}{x}+3x^2\,\ell n\,x\rightarrow\) Product Rule

\(=x^2+3x^2\,\ell n\,x\)

\(\dfrac {d^2y}{dx^2}=\dfrac {d}{dx}(x^2+3x^2\ell n\,x)\)

\(=2x+3\,\dfrac {d}{dx}(x^2\ell n\,x)\)

\(=2x+3\left [x^2×\dfrac {1}{x} +2x\;\ell n\,x\right]\)

\(= 2x+3x+6\,x\ell n\,x\)

\(= 5x+6\,x\ell n\,x\)

\(= x(5+6\,\ell n\,x)\)

If \(y=x^3\,\ell n\,x\), find \(\dfrac {d^2y}{dx^2}\).

A

\(x(5+6\,\ell n\,x)\)

.

B

\(x^2(6+5\,\ell n\,x)\)

C

\((sin\,x)(2+x^2)\)

D

\(\dfrac {\ell n\,x}{x}\)

Option A is Correct

Finding the Value of Parameter when Derivative at Some Value of x is given

  • Suppose  \(f(x)=Cx^2-\ell n\,x\)

If \(f'(1)=1\), then

 \(\Rightarrow2C\,x-\dfrac {1}{x}=1\)

\(\Rightarrow2C-1=1\)

\(\Rightarrow C=1\)

So, \(C=1\), when \(f'(1)=1\)

Illustration Questions

Let  \(f(x)=\ell n (tan\,x)+\alpha\,x\), find the value of \(\alpha\) if \(f'\left (\dfrac {\pi}{4}\right)=3\).

A \(\alpha =2\)

B \(\alpha =-2\)

C \(\alpha =5\)

D \(\alpha =1\)

×

\(f(x)=\ell n (tan\,x)+\alpha\,x\)

\(\Rightarrow f'(x)=\dfrac {1}{tan\,x}× sec^2\,x+\alpha\)

\(\Rightarrow f'\Big(\dfrac {\pi}{4}\Big)=\dfrac {sec^2 (\pi/4)}{tan\, (\pi/4)}+\alpha\)

\(=\dfrac {2}{1}+\alpha\)

\(=2+\alpha\)

\(\Rightarrow2+\alpha=3\)

\(\Rightarrow\alpha=1\)

Let  \(f(x)=\ell n (tan\,x)+\alpha\,x\), find the value of \(\alpha\) if \(f'\left (\dfrac {\pi}{4}\right)=3\).

A

\(\alpha =2\)

.

B

\(\alpha =-2\)

C

\(\alpha =5\)

D

\(\alpha =1\)

Option D is Correct

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