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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 :$$\dfrac {d}{dx}\left (\ell n (g(x)\right)=\dfrac {1}{g(x)}×g'(x)$$

#### 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 $$x=a$$.

#### 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

# 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.

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

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

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

#### 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.

#### 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

# 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 $$y-f(a)=f'(a)(x-a)$$

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

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

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

#### 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

# 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.

#### 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 $$f'(1)=1$$

#### 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