Informative line

### Problems On Combination Of Rules

Learn derivative chain rule and product & quotient rule combined problems. Practice equation differentiating composite function of functions use the chain quotient, product rule formulas.

# Chain Rule:

For differentiating composite function of two functions we use the following formula.

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

This is called Chain Rule

• In Leibniz Notation  if  $$y = f(x)$$ and $$u= g(x)$$
• $$\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$$

$$\dfrac{d}{dx} \underbrace f _{\text {outer function}} \,\,\underbrace {(g(x))}_{\text { inner function }} = \underbrace {f'}_{\text derivative\,of \,outer \,function} \,\,\,\underbrace {(g(x))}_ {\text {Inner function} } \,\,\,\underbrace {g'(x)} _{\text derivative\,of \,inner \,function}$$

e.g. If  $$h(x)= cos \,x^2$$  find $$h'(x)$$.

$$h(x) = f(g(x))$$ where $$g(x) = x^2$$ and $$f(x) = cos\,x$$

$$\therefore \,h'(x) = \underbrace{-sin}_ {\text derivative \,of \,cos.}×\overbrace{(x^2)}^{Inner\,function } × \underbrace {2x}_{\text derivative\, of\,inner\,function }$$

#### If $$y = tan (2\,x^2+x+3)$$ then find  $$\dfrac{dy}{dx}$$.

A $$(4x+1) \,sec^2(2x^2+x+3)$$

B $$(2\,x+3) \,(sec^2\,2x)$$

C $$(4x+7) \,(sec^2\,x)$$

D $$(x+5) \,sec^2(2x+x+3)$$

×

$$y = tan (2\,x^2+x+3) = f(g(x))$$

Where $$f(x) =tan \,x$$ and $$g(x) = 2\,x^2+x+3$$

$$\therefore \dfrac{dy}{dx} = f'(g(x)) × g'(x) \to$$ Chain Rule

$$= sec^2 (g(x)) × g'(x)$$

$$=sec^2 (2x^2+x+3) × \dfrac{d}{dx} (2x^2+x+3)$$

$$= sec^2 (2x^2+x+3)× (4x+1)$$

### If $$y = tan (2\,x^2+x+3)$$ then find  $$\dfrac{dy}{dx}$$.

A

$$(4x+1) \,sec^2(2x^2+x+3)$$

.

B

$$(2\,x+3) \,(sec^2\,2x)$$

C

$$(4x+7) \,(sec^2\,x)$$

D

$$(x+5) \,sec^2(2x+x+3)$$

Option A is Correct

# Combination of Ratio (Quotient), Power and Chain Rule

Many complicated functions will be differentiated by combination of

(1) Power rule

(2) Quotient rule

(3) Chain rule

(1) Power rule :- $$\dfrac{d}{dx}(f(x))^{n} = n (f(x))^{n-1}f'(x)$$

(2) Quotient rule: This rule is used to find the derivative of a quotient of two functions.

If $$f(x)$$ and $$g(x)$$ are two functions then the derivative of $$\dfrac{f(x)}{g(x)}$$ is given as

$$\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)\;=\;\dfrac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$$  (3) Chain rule:-  It is used to find the derivative of composite functions.

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

#### If $$y = \sqrt{\dfrac{x^2 + 1}{x^2 + 2}}$$ find $$\dfrac{dy}{dx}$$.

A $$\dfrac{x}{(x^2 + 1) (x^2 + 2)^2}$$

B $$\dfrac{x^2}{(x^2 + 1)(x^2 + 2)}$$

C $$5 \; sin \; x$$

D $$\dfrac{x}{(x^2 + 1)^{1/2}(x^2 + 2 )^{3/2}}$$

×

$$y= \left(\dfrac{x^2+1}{x^2 + 2}\right)^{1/2} = (f(x))^n$$   where $$f(x) = \dfrac{x^2 + 1}{x^2 + 2}\;$$ and $$n = 1/ 2$$

$$\dfrac{dy}{dx} = n (f(x))^{n-1} f'(x) = \dfrac{1}{2} \left(\dfrac{x^2 + 1}{x^2 + 2}\right)^{-1/2} × \dfrac{d}{dx} \left(\dfrac{x^2 + 1}{x^2 + 2}\right)$$  (Power rule)

$$=\dfrac{1}{2} \left(\dfrac{x^2 + 2}{x^2 + 1}\right)^{1/2} × \left(\dfrac{(x^2 + 2) \dfrac{d}{dx}(x^2 + 1) -(x^2+1)\dfrac{d}{dx}(x^2 + 2)} {(x^2 + 2)^2}\right)$$  (Quotient rule)

$$=\dfrac{1}{2} \left(\dfrac{x^2 + 2}{x^2 + 1}\right)^{1/2} × \dfrac{2x(x^2 + 2)-2x (x^2 + 1)}{(x^2 + 2)^2}$$

$$=\dfrac{1}{2} \left(\dfrac{x^2 + 2}{x^2 + 1}\right)^{1/2} × \dfrac{2x[(x^2 + 2)-(x^2 + 1)]}{(x^2 + 2)^2}$$

$$=\dfrac{1}{2} \left(\dfrac{x^2 + 2}{x^2 + 1}\right)^{1/2} × \dfrac{2x}{(x^2 +2)^2} = \dfrac{x}{(x^2 + 1)^{1/2} (x^2 + 2)^{3 /2}}$$

### If $$y = \sqrt{\dfrac{x^2 + 1}{x^2 + 2}}$$ find $$\dfrac{dy}{dx}$$.

A

$$\dfrac{x}{(x^2 + 1) (x^2 + 2)^2}$$

.

B

$$\dfrac{x^2}{(x^2 + 1)(x^2 + 2)}$$

C

$$5 \; sin \; x$$

D

$$\dfrac{x}{(x^2 + 1)^{1/2}(x^2 + 2 )^{3/2}}$$

Option D is Correct

# Derivative at Particular Values of x Using Chain Rule

• Let $$y= f(x)$$ be a given function ,then $$f'(a)$$ will mean the value of the derivative of the function at  $$x=a$$ . This means that we first differentiate the function with respect to $$x$$  and then put $$x=a$$ in the obtained expression.
• e.g. $$f(x) = sin\,2x$$

then $$f'\left(\dfrac{\pi}{4}\right) = 2\,cos\,2x \Bigg]_{x=\dfrac{\pi}{4}}$$

$$= 2× cos\dfrac{\pi}{2} = 2× 0=0.$$

#### If $$h(x) = f(g(f(2\,x)))$$  then find the value of $$h'(2)$$  if $$f(4) =3, \;g'(3)=7,\;g(3)=4$$ and $$f'(4)=-5$$.

A $$175$$

B $$-125$$

C $$80$$

D $$350$$

×

$$h(x) = f(g(f(2x)))$$

$$\Rightarrow h'(x) = f'(g(f(2x)))× g'(f(2x))× f'(2x)× 2\;\;\rightarrow\;Chain\;Rule$$

put  $$x=2$$  as  $$h'(2)$$ is desired .

$$\Rightarrow h'(2) = f'(g(f(4))) × g'(f(4)) × f'(4) × 2$$

$$=f'(g(3))\times\;g'(3)\times\,f'(4)\times\,2$$

$$= f'(4) × g'(3) × f'(4) × 2$$

$$= -5× 7× (-5)× 2=350$$

### If $$h(x) = f(g(f(2\,x)))$$  then find the value of $$h'(2)$$  if $$f(4) =3, \;g'(3)=7,\;g(3)=4$$ and $$f'(4)=-5$$.

A

$$175$$

.

B

$$-125$$

C

$$80$$

D

$$350$$

Option D is Correct

# Combination of Functions

Let us say, we have two functions $$'f'$$ and $$'g'$$, then we can make new functions from these by suitable combinations. Four such combinations are

(1) $$f + g$$        (2) $$f - g$$       (3) $$f g$$           (4) $$\dfrac {f}{g }$$

The definition is similar to the way we define addition, subtraction, division & multiplication in real numbers.

So

$$(f + g) (x) = f(x) + g(x)$$                    $$fg(x) = f(x)\; g(x)$$

$$(f – g) (x) = f(x) – g(x)$$                         $$\Big(\dfrac {f}{g}\Big)(x) = \Large \frac {f(x)}{g(x)}$$

#### Let $$f(x) = 5x^2 + 3$$ and $$g(x) = 2x+7$$ then find the value of  $$(f - g)(5)$$.

A 111

B 122

C 56

D 24

×

$$(f - g) (x) = f(x) - g(x)$$

In this case,

$$f(x)=5x^2+3,\;g(x)=2x+7$$

$$\therefore \;(f - g) (5) = f(5) - g(5)$$

$$= f(5×25+3) - (2×5+7)$$

$$= 128-17$$

$$=111$$

### Let $$f(x) = 5x^2 + 3$$ and $$g(x) = 2x+7$$ then find the value of  $$(f - g)(5)$$.

A

111

.

B

122

C

56

D

24

Option A is Correct

# Combination of Chain Rule and Power Rule

$$\dfrac{d}{dx}(f(x))^n = n(f(x))^{n-1}× f'(x)$$

• Derivative of $$n^{th}$$  power of any function is $$n$$ times $$(n-1)$$ powers of that function multiplied by derivative of the function.

Proof:

$$g(x) = (f(x))^n = \phi(h(x))$$

where $$\phi (x) = x^n$$   and $$h(x) = f(x)$$

$$\therefore g'(x) = \phi'(h(x)) × h'(x)$$

$$= n(f(x))^{n-1} × f'(x)$$

#### If $$y = (3x^2-4x+2)^{10}$$ then $$\dfrac{dy}{dx}=$$

A $$(60\,x-40)(3\,x^2-4\,x+2)^9$$

B $$(8\,x+3)(3\,x^2-4\,x+2)^{10}$$

C $$(4\,x-7)(3\,x^2-4\,x+2)^{10}$$

D $$14 \,sin\,x$$

×

$$y= (3x^2-4x+2)^{10} = (f(x))^n$$

where $$f(x) = 3x^2-4x+2$$ and $$n=10$$

$$\therefore \dfrac{dy}{dx} = n (f(x))^{n-1}× f'(x)$$

$$\dfrac{dy}{dx} =10\times(3x^2-4x+2)^9\times\dfrac{d\;}{dx}(3x^2-4x+2)$$

$$= 10× (3x^2-4x+2)^9 (6x-4)$$

$$= (60\,x-40)(3x^2-4x+2)^9$$

### If $$y = (3x^2-4x+2)^{10}$$ then $$\dfrac{dy}{dx}=$$

A

$$(60\,x-40)(3\,x^2-4\,x+2)^9$$

.

B

$$(8\,x+3)(3\,x^2-4\,x+2)^{10}$$

C

$$(4\,x-7)(3\,x^2-4\,x+2)^{10}$$

D

$$14 \,sin\,x$$

Option A is Correct

# Combination of Power Rule , Product Rule and Chain Rule

Many complicated functions will be differentiated by combination of

1. Power Rule

2. Product Rule

3. Chain Rule

1. Power Rule : $$\dfrac{d}{dx} (f(x))^n = n(f(x))^{n-1} f'(x)$$

2. Product Rule :  $$\dfrac{d}{dx} (f(x)g(x)) = f(x)\dfrac{d}{dx} g(x) + g(x)\dfrac{d}{dx} f(x)$$

3. Chain Rule : $$\dfrac{d}{dx} f(g(x)) = f'(g(x))\, g'(x)$$

#### If $$y = (x-2)^5 (2x+1)^4$$, find $$\dfrac{dy}{dx}$$.

A $$(x-2)^4 (2x+1)^3 (18x-11)$$

B $$(x+2) (18x+15)^4$$

C $$(x-2)^5 (2x+1)^4 (18x-11)$$

D $$(2x+1)^3 (x+2)^7$$

×

$$y = (x-2)^5 (2x+1)^4 = f(x)g(x)$$

where $$f(x) = (x-2)^5$$ and $$g(x) = (2x+1)^4$$

$$\dfrac{dy}{dx} = f(x) \,g'(x) + g(x) f'(x)\;\rightarrow\;Product\;Rule$$

$$= (x-2)^5\dfrac{d}{dx}(2x+1)^4 + (2x+1)^4\dfrac{d}{dx} (x-2)^5$$

$$= (x-2)^5 × 4(2x+1)^3 × 2 +(2x+1)^4 × 5(x-2)^4 × 1\;\rightarrow\;Power\;and\;Chain\;Rule$$

$$=8(x-2)^5 (2x+1)^3+5(2x+1)^4(x-2)^4$$

$$=(x-2)^4 (2x+1)^3[8(x-2)+5(2x+1)]$$

$$=(x-2)^4(2x+1)^3 (18x-11)$$

### If $$y = (x-2)^5 (2x+1)^4$$, find $$\dfrac{dy}{dx}$$.

A

$$(x-2)^4 (2x+1)^3 (18x-11)$$

.

B

$$(x+2) (18x+15)^4$$

C

$$(x-2)^5 (2x+1)^4 (18x-11)$$

D

$$(2x+1)^3 (x+2)^7$$

Option A is Correct