Informative line

### 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 $$\to \dfrac{d}{dx}(f(g(x)))=$$$$f'(g(x))$$×$$g'(x)$$

#### 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

# 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 $$f'(x)$$$$<0$$  $$\forall\,x\,\varepsilon(a,\,b)$$

#### 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  $$y=f(x)$$.

#### 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

# 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

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

$$=e^x$$

#### 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

# 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

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

#### 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.

#### 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

# 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.

#### 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. $$f'(x)$$$$=2e°sin\,0+e°cos\,0=1$$

$$\therefore$$ $$f'(0)$$$$=1$$

#### 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