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Calculus On Exponential Functions

Learn derivative of e^x, differentiation of exponential functions, Practice equation Integral of exponential function to find the absolute maximum or minimum values & tangent Line calculus.

Differentiation of Functions Involving Exponential Functions

To differentiate the function which involves exponential function. We use the following:

(1)  \(\dfrac{d}{dx}e^{f(x)}=e^{f(x)}× \)\(f'(x)\)

(2)  Product rule \(\to \dfrac{d}{dx}(f(x)\,g(x))=f(x)\)\(g'(x)\)+\(f'(x)\)\(g(x)\)

(3)  Chain rule \(\to \dfrac{d}{dx}(f(g(x)))=\)\(f'(g(x))\)×\(g'(x)\)

Illustration Questions

If \(f(x)=e^{-3x}sin2x\), find \(f'(x)\)

A \(e^{-3x}[5sin\,x+cos\,x]\)

B \(e^{5x}\)

C \(e^{-3x}[3cos\,2x-2sin\,2x]\)

D \(e^{-3x}[2cos\,2x-3sin\,2x]\)

×

\(f(x)=e^{-3x}sin\,2x\)

\(\Rightarrow\) \(f'(x)\)\(=\underbrace{e^{-3x}\dfrac{d}{dx}(sin\,2x)+sin\,2x\dfrac{d}{dx}(e^{-3x})}_\text{Product rule}\)

\(=\underbrace{e^{-3x}×2cos\,2x+sin\,2x×e^{-3x}×-3}_{Chain\,Rule}\)

\(=e^{-3x}[2cos\,2x-3sin\,2x]\)

If \(f(x)=e^{-3x}sin2x\), find \(f'(x)\)

A

\(e^{-3x}[5sin\,x+cos\,x]\)

.

B

\(e^{5x}\)

C

\(e^{-3x}[3cos\,2x-2sin\,2x]\)

D

\(e^{-3x}[2cos\,2x-3sin\,2x]\)

Option D is Correct

Equation of Tangent Line to the Curve having Exponential Function

  • Equation of tangent line at point \((a,\,f(a))\) of curve \(y=f(x)\) is given by

\(y-f(a)=\)\(f'(a)\)\((x-a)\)

Illustration Questions

Find the equation of tangent line to the curve \(y=\dfrac{e^{2x}}{x+1}\) at the point \((0,\,1)\).

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

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

C \(2x+y+3=0\)

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

×

Equation of tangent is  \(y-f(a)=\)\(f'(a)\)\((x-a)\)

\(f'(x)=\) \(\dfrac{dy}{dx}=\dfrac{(x+1)e^{2x}×2-e^{2x}×1}{(x+1)^2}\)

\(\therefore\) \(f'(0)\)\(=\dfrac{1×e°×2-e°}{(0+1)^2}=1\)

\(\therefore\) Equation of tangent at \((0,\,1)\) is 

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

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

Find the equation of tangent line to the curve \(y=\dfrac{e^{2x}}{x+1}\) at the point \((0,\,1)\).

A

\(x-y+1=0\)

.

B

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

C

\(2x+y+3=0\)

D

\(y-x-1=0\)

Option A is Correct

Finding Absolute Maximum and Absolute Minimum Values in an Interval

To find the absolute maximum or minimum values of a continuous function \(f\) in an interval [a,b], we do the following steps:

  1. Find all the critical points, i.e. solution to equation \(\dfrac{dy}{dx}=0\)  \(\left(\dfrac{d}{dx}\,e^x\right)=e^x\)
  2. The absolute maximum or minimum will occur either at critical point obtained in step 1 or at the end point.

Illustration Questions

Find the absolute maximum and minimum value of \(f(x)=x^2e^{-x}\) in the interval \([-1,\,3]\)

A Absolute maximum value \(=f(2)=\dfrac{4}{e^2}\), absolute minimum value \(=f(0)=0\)

B Absolute maximum value \(=f(-1)=e\), absolute minimum value \(=f(0)=0\)

C Absolute maximum value \(=f(2)=4e^2\), absolute minimum value \(=f(2)=\dfrac{4}{e^2}\)

D Absolute maximum value \(=f(3)=\dfrac{9}{e^3}\), absolute minimum value \(=f(4)=\dfrac{16}{e^4}\)  

×

Absolute maximum or minimum will occur either at critical points or end points of the interval.

\(f'(x)\)\(=0\)

\(\Rightarrow\,-x^2e^{-x}+2xe^{-x}=0\)

\(\Rightarrow\,e^{-x}[-x^2+2x]=0\)

\(\Rightarrow\,-x^2+2x=0\)

\(\Rightarrow\,-x(x-2)=0\)

\(\Rightarrow\,x=0,\;x=2\)

\(f(0)=0,\;f(-1)=e\)

\(f(2)=\dfrac{4}{e^2},\;f(3)=\dfrac{9}{e^3}\)

The greatest \(=f(-1)=e=\) Absolute maximum value

parallel least \(=f(0)=0=\) Absolute minimum value

Find the absolute maximum and minimum value of \(f(x)=x^2e^{-x}\) in the interval \([-1,\,3]\)

A

Absolute maximum value \(=f(2)=\dfrac{4}{e^2}\), absolute minimum value \(=f(0)=0\)

.

B

Absolute maximum value \(=f(-1)=e\), absolute minimum value \(=f(0)=0\)

C

Absolute maximum value \(=f(2)=4e^2\), absolute minimum value \(=f(2)=\dfrac{4}{e^2}\)

D

Absolute maximum value \(=f(3)=\dfrac{9}{e^3}\), absolute minimum value \(=f(4)=\dfrac{16}{e^4}\)

 

Option B is Correct

Intervals of Increase and Decrease for the Expression containing Exponential Function

  • A function is increasing in \((a,\,b)\) if \(f'(x)\)\(>0\;\forall\)  \(x\,\varepsilon(a,\,b)\)
  • A function is decreasing in \((a,\,b)\) if \(f'(x)\)\(<0\)  \(\forall\,x\,\varepsilon(a,\,b)\)

Illustration Questions

Find the intervals of increase and decrease of \(f(x)=\dfrac{1}{xe^x}\)

A \(f\) is increasing in \((-\infty,\,-1)\) and decreasing in \((-1,\,0)\,\cup(0,\,\infty)\)

B \(f\) is increasing in \((-\infty,\,2)\) and decreasing in \((2,\,\infty)\)

C \(f\) is increasing in \((-\infty,\,0)\) and decreasing in \((0,\,\infty)\)

D \(f\) is increasing in \((-\infty,\,-3)\) and decreasing in \((-3,\,0)\,\cup(0,\,\infty)\)

×

\(f(x)=\dfrac{1}{xe^{x}}\to\) domain is \(R-\{0\}\)

\(f'(x)\)\(=\dfrac{-(xe^x+e^x)}{(xe^x)^2}\)

\(=\dfrac{-e^x(x+1)}{x^2e^{2x}}\)

\(f'(x)\)\(>0\),

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

\(\Rightarrow\,(x+1)<0\)

\(\Rightarrow\,x<-1\)

\(\Rightarrow\,f\) is increasing in \((-\infty,\,-1)\)

\(f'(x)\)\(<0\),

\(\Rightarrow\,x\,\varepsilon(-1,\,0)\,\cup(0,\,\infty)\)

\(\therefore\) f is decreasing in \((-1,\,0)\;\cup(0,\,\infty)\)

Find the intervals of increase and decrease of \(f(x)=\dfrac{1}{xe^x}\)

A

\(f\) is increasing in \((-\infty,\,-1)\) and decreasing in \((-1,\,0)\,\cup(0,\,\infty)\)

.

B

\(f\) is increasing in \((-\infty,\,2)\) and decreasing in \((2,\,\infty)\)

C

\(f\) is increasing in \((-\infty,\,0)\) and decreasing in \((0,\,\infty)\)

D

\(f\) is increasing in \((-\infty,\,-3)\) and decreasing in \((-3,\,0)\,\cup(0,\,\infty)\)

Option A is Correct

Integral of Exponential Function

\(\dfrac{d}{dx}(e^x)=e^x\)

\(\Rightarrow\int e^xdx=e^x+C\)

If the integral function is of the form \(\displaystyle I=\int \)\(f'(x)\)\(e^{f(x)}dx\) then we make the substitution .

\(f(x)=t\Rightarrow f'(x)dx=dt\)

  • So the integral gets transformed to 

\(\displaystyle I=\int e^tdt=e^t+C\)

\(\therefore\,I=e^{f(x)}+C\)

\(\displaystyle\therefore\,\int\,\)\(f'(x)\)\(e^{f(x)}dx=e^{f(x)}+C\)

So whenever we see \(e^{f(x)}\) term we think of the substitution  \(y=f(x)\).

Illustration Questions

Find \(\displaystyle\,I=\int sin\,x×e^{cos\,x}dx\).

A \(-e^{cos\,x}+C\)

B \(e^{x^2}+C\)

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

D \(e^x+C\)

×

\(\displaystyle\,I=\int sin\,x×e^{cos\,x}dx\)

Put,  \(cos\,x=t\)

\(\Rightarrow\,-sin\,x\,dx=dt\)

\(\Rightarrow\,sin\,x\,dx=-dt\)

\(\displaystyle\,I=\int -e^tdt\)

\(=-e^t+C\)

\(=-e^{cos\,x}+C\)

Find \(\displaystyle\,I=\int sin\,x×e^{cos\,x}dx\).

A

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

.

B

\(e^{x^2}+C\)

C

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

D

\(e^x+C\)

Option A is Correct

Finding value of the Parameter of Exponential Functions satisfying Differential Equation

  • Sometimes an exponential function which involves a parameter will be given, that satisfies a particular differential equation.
  • Put the function and its derivative in that equation to obtain an equation in which the parameter is involved, solve for the parameter.

Illustration Questions

If \(f(x)=e^{\alpha x}\) (where \(\alpha\) is a constant) satisfies the equation \(f''(x)\)\(-5\)\(f'(x)\)\(+6f(x)=0,\) find the values of \(\alpha\).

A \(\alpha=1,\;\alpha=-2\)

B \(\alpha=5,\;\alpha=1\)

C \(\alpha=2,\;\alpha=3\)

D \(\alpha=6,\;\alpha=-2\)

×

\(f(x)=e^{\alpha x}\)

\(\Rightarrow\) \(f'(x)\)\(=\alpha \,e^{\alpha x}\)

\(\Rightarrow\) \(f''(x)\)\(=\alpha^2e^{\alpha x}\)

\(\therefore\) \(f''(x)\)\(-5\)\(f'(x)\)\(+6f(x)=0\)

\(\Rightarrow\,\alpha^2e^{\alpha x}-5\alpha\, e^{\alpha x}+6e^{\alpha x}=0\)

\(\Rightarrow\,\underbrace{e^{\alpha x}}_{\text{always>0}}[\alpha^2-5\alpha+6]=0\)

\(\Rightarrow\,\alpha^2-5\alpha+6=0\)

\(\Rightarrow\,(\alpha-2)(\alpha-3)=0\)

\(\Rightarrow\,\alpha=2,\,\alpha=3\)

If \(f(x)=e^{\alpha x}\) (where \(\alpha\) is a constant) satisfies the equation \(f''(x)\)\(-5\)\(f'(x)\)\(+6f(x)=0,\) find the values of \(\alpha\).

A

\(\alpha=1,\;\alpha=-2\)

.

B

\(\alpha=5,\;\alpha=1\)

C

\(\alpha=2,\;\alpha=3\)

D

\(\alpha=6,\;\alpha=-2\)

Option C is Correct

Derivatives at Particular Values of \(x\)

  • To obtain the derivative of function at particular values of \(x\), we first differentiate the function using rules of differentiation and then, put the value of \(x\)where the derivative is desired.
  • e.g. If \(f(x)=e^{2x}sin\,x\), then for \(f'(0)\)
  1. \(f'(x)\)\(=2e^{2x}sin\,x+e^{2x}cos\,x\)
  2. \(f'(x)\)\(=2e°sin\,0+e°cos\,0=1\)

\(\therefore\) \(f'(0)\)\(=1\)

Illustration Questions

If \(g(x)=e^{2x}f(x)\), find the value of \(g''(2)\) . If \(f(2)=1,\) \(f'(2)\)\(=-1,\) \(f''(2)\) \(=2\).

A \(2e^4\)

B \(2e^3\)

C \(\dfrac{3}{e}\)

D \(\dfrac{1}{e^2}\)

×

\(g(x)=e^{2x}f(x)\)

\(\Rightarrow\) \(g'(x)\) \(=e^{2x}\)\(f'(x)\)\(+2e^{2x}f(x)\)

\(\Rightarrow\,\)\(g''(x)\)\(=\big(e^{2x}\)\(f''(x)\)\(+2e^{2x}\)\(f'(x)\big)\)\(+\big(2e^{2x}\)\(f'(x)\)\(+4e^{2x}f(x)\big)\)

\(\therefore\) \(g''(x)\)\(=e^{2x}[\)\(f''(x)\)\(+4\)\(f'(x)\)\(+4f(x)]\)

\(\Rightarrow\) \(g''(2)\) \(=e^{4}\big[\)\(f''(2)\)\(+4\)\(f'(2)\)\(+4f(2)\big]\)

\(=e^4\big[2+(-4)+4\big]\)

\(=2e^4\)

If \(g(x)=e^{2x}f(x)\), find the value of \(g''(2)\) . If \(f(2)=1,\) \(f'(2)\)\(=-1,\) \(f''(2)\) \(=2\).

A

\(2e^4\)

.

B

\(2e^3\)

C

\(\dfrac{3}{e}\)

D

\(\dfrac{1}{e^2}\)

Option A is Correct

Derivative of Exponential Functions(\(e^x\))

\(e\) is a number such that  \(\lim\limits_{h\rightarrow0} \left ( \dfrac {e^h-1}{h} \right)=1\)

Consider,

\(\dfrac {d}{dx}(a^x)=\lim\limits_{h\rightarrow0} \left ( \dfrac {a^{x+h}-a^x}{h} \right)\)

\(=a^x\;\lim\limits_{h\rightarrow0} \left ( \dfrac {a^{h}-1}{h} \right)\)

\(\dfrac {d}{dx}(a^x)=a^x\;×\ell na \)

              \( =a^x\;\ell na\)

If, \(a=e\) then we say that 

\(\dfrac {d}{dx}\;e^x=e^x\,ln(e)\)

              \(=e^x\)

Illustration Questions

If \(f(x)=e^x\), find \(f'(x)\)

A \(2\,e^{2\,x}\)

B \(e^{x}\)

C \(4\,e^{2\,x}\)

D \(2\,e^{x}\)

×

\(\dfrac {d}{dx}(a^x)=\lim\limits_{h\rightarrow0} \left ( \dfrac {a^{x+h}-a^x}{h} \right)\)

\(=a^x\;\lim\limits_{h\rightarrow0} \left ( \dfrac {a^{h}-1}{h} \right)\)

\(\dfrac {d}{dx}(a^x)=a^x\;×\ell na \)

              \( =a^x\;\ell na\)

If \(a=e\) then, we say that 

\(\dfrac {d}{dx}\;e^x=e^x\,\,ln(e)\)

              \(=e^x\)

If \(f(x)=e^x\), find \(f'(x)\)

A

\(2\,e^{2\,x}\)

.

B

\(e^{x}\)

C

\(4\,e^{2\,x}\)

D

\(2\,e^{x}\)

Option B is Correct

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