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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.
  • \(\dfrac {d}{dx}\,e^{g(x)}=e^{g(x)}\,g'(x)\)

Illustration Questions

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.

Illustration Questions

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

Illustration Questions

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.

Illustration Questions

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  \(\displaystyle \int\dfrac {1}{x}\;dx=\ell n\,|\,x\,|+C\)

Illustration Questions

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  \(g(x) = t\).

Illustration Questions

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. 

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

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

\(=5[\ell n\,3-\ell n\,2]\)

\(=5\;\ell n\;\dfrac {3}{2}\)

Illustration Questions

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 \(\dfrac {d^2y}{dx^2}<0\).

Illustration Questions

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

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