Informative line

### More Calculus On Log Function

Learn derivative of log x function & absolute maximum & minimum values of expression which have exponential function. Practice concavity of expressions which involve logarithmic function & Evaluating by Appropriate Substitution.

# Absolute Maximum and Minimum Values of Expression which have Exponential Function

• To find the local maximum or minimum values of a function $$f$$which contain exponential function, we put $$f'(x)=0$$ and solve for the values of $$x$$(say $$\alpha_1\,,\alpha_2\,,...\alpha_n\,$$).
• If$$f '' (\alpha _i) > 0$$, then we say that  $$x = \alpha_i$$ is a local minima.
• If  $$f '' (\alpha _i) < 0$$, then we say that  $$x = \alpha_i$$ is a local maxima.
• $$\dfrac {d}{dx}\,e^{g(x)}=e^{g(x)}\,g'(x)$$

#### Find the values of  $$x$$ for which $$f(x)=e^{4x}-2e^x$$ take local maximum or local minimum values .

A Local minimum value of $$x = \dfrac {1}{3}\,\ell n\dfrac {1}{2}$$, no local maxima

B Local minimum value of $$x=\ell n\,2$$, no local maxima

C Local minimum value of $$x =\ell n\dfrac {1}{2}$$, no local maxima

D Local maximum value of $$x = \ell n\,2$$, no local minima

×

For local maxima or minima $$f'(x)=0$$

$$f(x)=e^{4x}-2e^{x}$$

$$\Rightarrow f'(x)=4e^{4x}-2e^{x}$$

$$\therefore\;f'(x)=0$$

$$\Rightarrow4e^{4x}-2e^{x}=0$$

$$\Rightarrow 2e^x[2e^{3x}-1]=0$$

$$\Rightarrow e^{3x}=\dfrac {1}{2}$$

$$\Rightarrow x=\dfrac {1}{3}\ell n \Big(\dfrac{1}{2}\Big)$$

$$\therefore\;f''(x)=16e^{4x}-2e^{x}$$

$$f''\left (\dfrac {1}{3}\ell n \Big(\dfrac{1}{2}\Big)\right)= 16\,e^{\left (\dfrac {4}{3}\ell n \dfrac{1}{2}\right)}- 2e^{\left (\dfrac {1}{3}\ell n \dfrac{1}{2}\right)}$$

$$=16×\Big(\dfrac{1}{2}\Big)^{4/3}-2×\Big(\dfrac{1}{2}\Big)^{1/3}>0$$

$$\therefore \;x=\dfrac {1}{3}\,\ell n\dfrac {1}{2}$$ is a point of local minima , no local maxima.

### Find the values of  $$x$$ for which $$f(x)=e^{4x}-2e^x$$ take local maximum or local minimum values .

A

Local minimum value of $$x = \dfrac {1}{3}\,\ell n\dfrac {1}{2}$$, no local maxima

.

B

Local minimum value of $$x=\ell n\,2$$, no local maxima

C

Local minimum value of $$x =\ell n\dfrac {1}{2}$$, no local maxima

D

Local maximum value of $$x = \ell n\,2$$, no local minima

Option A is Correct

# Concavity of the Exponential Curve

• A curve $$y = f(x)$$ is concave upward, if $$f'' (x) > 0$$ for given interval.
• A curve $$y = f(x)$$ is concave downward, if $$f'' (x) < 0$$ for given interval.

#### On what interval is the curve $$y=4e^x-e^{-2x}$$ concave upwards?

A $$(0,\infty)$$

B $$(2,\infty)$$

C $$(-\infty, 3)\cup(5,\infty)$$

D $$(-\infty, 6)$$

×

For intervals where function is concave upward, $$f'' (x) > 0$$

$$f(x)=4e^{x}-e^{-2x}$$

$$\Rightarrow f'(x)=4e^x+2e^{-2x}$$

$$\Rightarrow f'' (x) =4e^{x}-4e^{-2x}$$

$$f''(x)>0$$

$$\Rightarrow4e^{x}-4e^{-2x}>0$$

$$\Rightarrow 4[e^{x}-e^{-2x}]>0\;$$

$$\Rightarrow e^x>e^{-2x}$$

$$\Rightarrow e^{x}>\dfrac {1}{e^{2x}}\;$$

$$\Rightarrow e^{3x}>1 \;$$

$$\Rightarrow 3x> \ell n\,1$$

$$\Rightarrow x>0$$

$$f$$ is concave upwards in $$(0,\infty)$$.

### On what interval is the curve $$y=4e^x-e^{-2x}$$ concave upwards?

A

$$(0,\infty)$$

.

B

$$(2,\infty)$$

C

$$(-\infty, 3)\cup(5,\infty)$$

D

$$(-\infty, 6)$$

Option A is Correct

# Derivative of General Exponential Function

• If $$f(x)=a^x$$ (where $$a>0$$), then $$f'(x)=a^x\,\ell n\,a$$
• This means that $$\dfrac {d}{dx}(a^x)=a^x\,\ell n\,a$$

e.g. $$\dfrac {d}{dx}(2^x)=2^x\,\ell n\,2$$

• If we take $$a=e$$, then $$f'(x)=e^x\,\ell n\,e=e^x$$, which is already known.$$\dfrac {d}{dx}(a^{f(x)})=a^{f(x)}\ell n\,a×f'(x)$$

Combine the above rule with the Chain Rule.

• e.g. $$\dfrac {d}{dx}(5^{x^3})=5^{x^3}\ell n\,5×3x^2=3\,\ell n\,5\;x^2×5^{x^3}$$

#### If $$f(x)=3^{(2sin\,x+cos\,x)}$$, find $$f'(x)$$.

A $$e^{(2sin\,x+cos\,x)}-\ell n\,3$$

B $$(2sin\,x+cos\,x)×3^{(2\,sin\,x+cos\,x)}$$

C $$5sin^2\,x+cos^3\,x$$

D $$(2cos\,x-sin\,x)×3^{(2\,sin\,x+cos\,x)}×\ell n\,3$$

×

$$f(x)=3^{(2sin\,x+cos\,x)}$$

$$\Rightarrow f'(x)=3^{(2sin\,x+cos\,x)}×\ell n\,3×\dfrac {d}{dx}(2sin\,x+cos\,x)$$

$$\Rightarrow f'(x)=3^{(2sin\,x+cos\,x)}×\ell n\,3×(2cos\,x-sin\,x)$$

$$=(2cos\,x-sin\,x)×3^{(2\,sin\,x+cos\,x)}×\ell n\,3$$

### If $$f(x)=3^{(2sin\,x+cos\,x)}$$, find $$f'(x)$$.

A

$$e^{(2sin\,x+cos\,x)}-\ell n\,3$$

.

B

$$(2sin\,x+cos\,x)×3^{(2\,sin\,x+cos\,x)}$$

C

$$5sin^2\,x+cos^3\,x$$

D

$$(2cos\,x-sin\,x)×3^{(2\,sin\,x+cos\,x)}×\ell n\,3$$

Option D is Correct

# Integration of a General Exponential Function

• $$\int a^x\;dx=\dfrac {a^x}{\ell n\,a}+C\;\;\; (a\neq1)$$
• If we put $$a=e$$ we get,  $$\int e^x\;dx=e^x+C$$    which is already known.

#### Find the value of  $$I=\int\limits_1^5 3^x\;dx$$.

A $$\dfrac {242}{\ell n\,3}$$

B $$\dfrac {\ell n\,5}{\ell n\,2}$$

C $$\dfrac {512}{\ell n\,3}$$

D $$5$$

×

$$\int\, 3^x\;dx=\dfrac {3^x}{\ell n \,3}+C$$

$$\therefore \;I=\int\limits_1^5\, 3^x\;dx$$

$$=\Bigg[\dfrac {3^x}{\ell n \,3}\Bigg]_1^5$$

$$=\dfrac {1}{\ell n\,3}\Big[3^5-3^1\Big]$$

$$=\dfrac {243-1}{\ell n\,3}$$

$$=\dfrac {242}{\ell n\,3}$$

### Find the value of  $$I=\int\limits_1^5 3^x\;dx$$.

A

$$\dfrac {242}{\ell n\,3}$$

.

B

$$\dfrac {\ell n\,5}{\ell n\,2}$$

C

$$\dfrac {512}{\ell n\,3}$$

D

$$5$$

Option A is Correct

# Integral having Result as Logarithmic Function

• Consider $$\dfrac {d}{dx}(\ell n\,x)=\dfrac {1}{x}$$ ....(1)

$$\dfrac {d}{dx}(\ell n\,(-x))=\dfrac {1}{-x}×(-1)=\dfrac {1}{x}$$ ...(2)

• By (1) and (2), we can say that  $$\displaystyle \int\dfrac {1}{x}\;dx=\ell n\,|\,x\,|+C$$

#### Find $$I=\displaystyle\int\dfrac {1}{(2x+3)}\;dx$$.

A $$\ell n\,x+tan\,x+C$$

B $$\dfrac {sin\,x}{x}+C$$

C $$\dfrac {1}{2}\ell n\,|\,2x+3\,|+C$$

D $$\dfrac {1}{3}\ell n\,|\,3x+4\,|+C$$

×

$$I=\displaystyle\int\dfrac {1}{(2x+3)}\;dx$$

Put  $$2x+3=t$$,

$$\Rightarrow 2\,dx=dt$$

$$\Rightarrow dx=\dfrac {dt}{2}$$

$$\Rightarrow I=\displaystyle\int\dfrac {1}{2}\;\dfrac {dt}{t}$$

$$\Rightarrow I=\dfrac {1}{2}\;\ell n|\,t\,|+C$$

$$=\underbrace{\dfrac {1}{2}\;\ell n|\,2x+3\,|}_{go\, to\, original\, variable\,x}+C$$

### Find $$I=\displaystyle\int\dfrac {1}{(2x+3)}\;dx$$.

A

$$\ell n\,x+tan\,x+C$$

.

B

$$\dfrac {sin\,x}{x}+C$$

C

$$\dfrac {1}{2}\ell n\,|\,2x+3\,|+C$$

D

$$\dfrac {1}{3}\ell n\,|\,3x+4\,|+C$$

Option C is Correct

# Evaluating Integrals which lead to Integration of 1/x by Appropriate Substitution

• If we make an appropriate substitution in some integrals, it is converted to the form $$\displaystyle\int \dfrac {1}{x}\;dx$$ which can be written as $$\ell n(x)+C$$.

Suppose we have to find,

$$I=\displaystyle\int \dfrac {cos\,x}{sin\,x}\;dx$$

We put  $$sin\,x=t$$

$$\Rightarrow cos\,x\;dx = dt$$

$$\Rightarrow I=\displaystyle\int \dfrac {dt}{t}=\ell n\,|t|+C$$

$$\Rightarrow \ell n\, | sin\,x|+C$$

Usually integral of the form

$$I=\displaystyle\int \dfrac {f'(x)}{f(x)}\;dx$$

gets converted to this form.

e.g.

$$I=\displaystyle\int \dfrac {cos\,x}{1+sin\,x}\;dx$$

Put  $$1+sin\,x=t$$

$$\Rightarrow cos\,x\;dx=dt$$

$$I=\displaystyle\int \dfrac {dt}{t}\;=\ell n (t)+C$$

$$\Rightarrow \ell n|1+sin\,x|+C$$

• Most of the times these integrals are of the form $$\displaystyle\int \dfrac {g'(x)}{g(x)}\;dx$$ ,where $$g(x)$$ is an expression. So, whenever we come across such a pattern of integrand we put  $$g(x) = t$$.

#### Find $$I=\displaystyle\int \dfrac {e^x}{2e^x+1}\;dx$$.

A $$2\ell n\, |cos\,x|+C$$

B $$\dfrac {1}{2}\ell n\, |2e^x+1|+C$$

C $$\ell n\, |sin\,x|+C$$

D $$\dfrac {1}{3}\ell n\, |e^x+2|+C$$

×

$$I=\displaystyle\int \dfrac {e^x}{2e^x+1}\;dx$$

Put  $$2e^x+1=t$$

$$\Rightarrow 2e^xdx=dt$$

$$\Rightarrow e^x\;dx=\dfrac {1}{2}\;dt$$

$$\Rightarrow I=\dfrac {1}{2}\displaystyle\int \dfrac {dt}{t}$$

$$\therefore \;I=\dfrac {1}{2}\;\ell n (t)+C$$

$$=\underbrace{\dfrac {1}{2}\,\ell n |2e^x+1|}_{go \,to\,the \,original\,variable\,x} +C$$

### Find $$I=\displaystyle\int \dfrac {e^x}{2e^x+1}\;dx$$.

A

$$2\ell n\, |cos\,x|+C$$

.

B

$$\dfrac {1}{2}\ell n\, |2e^x+1|+C$$

C

$$\ell n\, |sin\,x|+C$$

D

$$\dfrac {1}{3}\ell n\, |e^x+2|+C$$

Option B is Correct

# Definite Integrals which Involve Integration of 1/x

• Use Fundamental Theorem of Calculus after evaluating the indefinite integral of the integrand.
• e.g.

$$I=\displaystyle\int\limits_2^3\dfrac {5}{x}\;dx$$

$$=5\,\Big[\ell n\,x\Big]_2^3$$

$$=5[\ell n\,3-\ell n\,2]$$

$$=5\;\ell n\;\dfrac {3}{2}$$

#### Find the value of  $$I=\displaystyle\int\limits_2^3\dfrac {dx}{(2x-1)}$$.

A $$\dfrac {1}{2}\;\ell n\;\dfrac {5}{3}$$

B $$\dfrac {1}{3}\;\ell n\;\dfrac {4}{3}$$

C $$5\;\ell n\;2$$

D $$-9$$

×

$$I=\displaystyle\int\limits_2^3\dfrac {dx}{(2x-1)}$$

Put  $$2x-1=t$$

$$\Rightarrow 2\;dx=dt$$

$$\Rightarrow dx=\dfrac {dt}{2}$$

Also, when $$x=3\rightarrow t = 5 \\ x=2\rightarrow t = 3$$

$$\therefore$$ $$I=\displaystyle\int\limits_3^5\dfrac {1}{2}\;\dfrac {dt}{t}$$

$$\Rightarrow I=\dfrac {1}{2}\;\Big[\ell n\,t\Big]_3^5$$

$$=\dfrac {1}{2}[\ell n\,5-\ell n\,3]$$

$$=\dfrac {1}{2}\;\ell n\;\dfrac {5}{3}$$

### Find the value of  $$I=\displaystyle\int\limits_2^3\dfrac {dx}{(2x-1)}$$.

A

$$\dfrac {1}{2}\;\ell n\;\dfrac {5}{3}$$

.

B

$$\dfrac {1}{3}\;\ell n\;\dfrac {4}{3}$$

C

$$5\;\ell n\;2$$

D

$$-9$$

Option A is Correct

# Concavity of Expressions which Involve Logarithmic Function

A curve $$y = f(x)$$ is concave upwards, when $$\dfrac {d^2y}{dx^2}>0$$ and concave downwards, when $$\dfrac {d^2y}{dx^2}<0$$.

#### Find the interval in which the graph of the curve  $$y=2x^2+\ell n\,x$$ is concave upwards.

A $$x\in\left (\dfrac {1}{2}, \,5\right)$$

B $$x\in\left (\dfrac {1}{2}, \,\infty\right)$$

C $$x\in\left (-\infty, \,4\right)$$

D $$x\in\left (1, \,7\right)$$

×

$$y=2x^2+\ell n\,x$$

$$\dfrac {dy}{dx}=4x+\dfrac {1}{x}$$

$$\dfrac {d^2y}{d^2x}=4-\dfrac {1}{x^2}$$

For concave upwards, $$\dfrac {d^2y}{dx^2}>0$$

$$\Rightarrow 4-\dfrac {1}{x^2}>0$$

$$\Rightarrow \dfrac {4x^2-1}{x^2}>0$$

$$\Rightarrow \dfrac {(2x-1)(2x+1)}{x^2}>0$$

$$\Rightarrow (2x-1)>0$$

$$2x+1$$ and $$x^2$$ are positive as $$x$$ must be positive for $$\ell n\,x$$ to be defined).

$$\Rightarrow x> \dfrac {1}{2}$$

$$\Rightarrow x\in \left ( \dfrac {1}{2},\,\infty \right)$$

### Find the interval in which the graph of the curve  $$y=2x^2+\ell n\,x$$ is concave upwards.

A

$$x\in\left (\dfrac {1}{2}, \,5\right)$$

.

B

$$x\in\left (\dfrac {1}{2}, \,\infty\right)$$

C

$$x\in\left (-\infty, \,4\right)$$

D

$$x\in\left (1, \,7\right)$$

Option B is Correct