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

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 \(f(x) = cos\,x\)

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

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

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

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

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

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.

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

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

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

\(= 2× cos\dfrac{\pi}{2} = 2× 0=0.\)

A \(175\)

B \(-125\)

C \(80\)

D \(350\)

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

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

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

A 111

B 122

C 56

D 24

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

- 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 \(h(x) = f(x)\)

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

\(= n(f(x))^{n-1} × f'(x)\)

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 \(14 \,sin\,x\)

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

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