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### Derivative Of Trigonometric Functions

Practice problems derivative of trigonometric functions & Chain rule examples, Learn Derivatives of trig functions.

# Derivative of Sine Function(sin x)

$$\dfrac{d}{dx}(\sin \; x)\;=\;\cos\; x$$     ( where $$x$$ is in radians )

Graph of sin x is as shown.

The derivative of $$\sin\;x$$ with respect to $$x$$ is $$\cos \; x$$

This means that the slope of the tangent to the graph of $$y = \sin \; x$$  curve at any point $$(x, \; \sin \; x)$$ on it, is $$\cos \; x$$.

Thus the graph for cos x will be as shown.

• Consider an example to understand this.

$$y=x^4\sin x$$

Since the above function is the product of two functions: one algebraic and the other trigonometric, so to find the derivative we will use product rule.

$$\dfrac{dy}{dx}=\dfrac{d}{dx}(x^4\sin x)$$

Using product rule-

$$\dfrac{dy}{dx}=\sin x\;\dfrac{d}{dx}(x^4)+x^4\dfrac{d}{dx}(\sin x)$$

$$\dfrac{dy}{dx}=\sin x\,(4x^3)+x^4\,\cos x$$

$$\dfrac{dy}{dx}=4x^3\,\sin x+x^4\,\cos x$$

#### If   $$y = 2 x ^3\;sin\;x$$  find  $$\dfrac{dy}{dx}$$.

A $$7 x \;sin \;x + 4 x^2 \;cos \; x$$

B $$2x^3\; cos \; x + 6 x ^2 \; sin \; x$$

C $$4x^2\; sin \;x - 2x\; cos \; x$$

D $$5x^2\; sin \; x + 2 x\; cos \;x$$

×

$$y = 2 x^3 \; sin \; x$$

$$\Rightarrow\dfrac{dy}{dx}\;=\; \dfrac{d}{dx}(2x^3\; sin\; x)$$

$$2\dfrac{d}{dx}(x^3\;sin \; x)\;=\underbrace{\;2\left[x^3 \dfrac{d}{dx}\;sin\; x + sin \; x \dfrac{d}{dx}x^3\right]}_\text{Product rule }$$

$$= 2 \left[x^3\; cos \; x + sin \; x × 3x^2\right]$$

$$=\,2x^3\; cos \; x + 6 x^2 \; sin \; x$$

### If   $$y = 2 x ^3\;sin\;x$$  find  $$\dfrac{dy}{dx}$$.

A

$$7 x \;sin \;x + 4 x^2 \;cos \; x$$

.

B

$$2x^3\; cos \; x + 6 x ^2 \; sin \; x$$

C

$$4x^2\; sin \;x - 2x\; cos \; x$$

D

$$5x^2\; sin \; x + 2 x\; cos \;x$$

Option B is Correct

# Derivative of Cosine Function (cos x)

$$\dfrac{d}{dx}\; (\cos \; x) =\,- \sin \; x$$    ( where $$x$$ is in radian )

The derivative of cos $$x$$ with respect to $$x$$ is $$-\sin\,x$$

This means that the slope of tangent to the graph of $$y = \cos \,x$$ at any point ($$x$$,  $$\cos\,x$$ ) on it, is $$-\sin\,x$$

• Consider an example to understand this.

$$y=\dfrac{5\cos x}{x}$$

To find the derivative

$$\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{5\cos x}{x}\right)$$    [By using quotient rule]

$$\dfrac{dy}{dx}=\dfrac{x\frac{d}{dx}(5\cos x)-5\cos x\frac{d}{dx}(x)}{x^2}$$

$$\dfrac{dy}{dx}=\dfrac{-x\,5\sin x-5\,\cos x}{x^2}$$

$$\dfrac{dy}{dx}=\dfrac{-5x\,\sin x-5\,\cos x}{x^2}$$

#### If $$y = \dfrac{2 \;cos \; x}{x^2}$$ find  $$\dfrac{dy}{dx}$$.

A $$\dfrac{2 sin \; x}{4}+\dfrac{ 4 cos\; x}{x^2}$$

B $$4x^2sin\; x + 5 x \; cos \;x$$

C $$8 sin \; x - 2 cos \; x$$

D $$\dfrac{-2 sin \; x}{x^2}+ \dfrac{4 cos \; x}{x^3}$$

×

$$y = \dfrac{2 cos \; x}{x^2}$$   then  $$\dfrac{dy}{dx} = \dfrac{d}{dx}\left(\dfrac{2 cos \; x}{x^2}\right)$$

$$=\;\;2\dfrac{d}{dx}\left(\dfrac{cos \; x}{x^2}\right) = \underbrace {2 \left[\dfrac{x^2 \dfrac{d}{dx} (cos \; x) - cos\,x \dfrac{d}{dx} x^2}{x^4}\right]}_\text{Quotient rule}$$

$$=\;\; 2 \left[ \dfrac{-x^2 \; sin \; x - 2 x\; cos \; x}{x^4}\right]$$

$$=\;\;\dfrac{-2 sin \; x}{x^2}- \dfrac{4 cos \; x}{x^3}$$

### If $$y = \dfrac{2 \;cos \; x}{x^2}$$ find  $$\dfrac{dy}{dx}$$.

A

$$\dfrac{2 sin \; x}{4}+\dfrac{ 4 cos\; x}{x^2}$$

.

B

$$4x^2sin\; x + 5 x \; cos \;x$$

C

$$8 sin \; x - 2 cos \; x$$

D

$$\dfrac{-2 sin \; x}{x^2}+ \dfrac{4 cos \; x}{x^3}$$

Option D is Correct

# Derivative of secant Function(sec x)

• $$\dfrac {d}{dx}(\sec\,x)=\sec\,x\,\tan\,x$$     (where $$x$$ is in radians)

The derivative of  $$\sec\,x$$ with respect to  $$x$$ is  $$\sec\,x\,\tan\,x$$

Proof:

$$\dfrac {d}{dx}(\sec\,x)=\dfrac {d}{dx}\left (\dfrac {1}{\cos\,x}\right)$$

$$=\dfrac {\cos\,x\,\dfrac {d}{dx}1-1×\dfrac {d}{dx}\,\cos\,x}{\cos^2\,x}$$

$$=\dfrac {0+\sin\,x}{\cos^2\,x}$$

$$=\dfrac {\sin\,x}{\cos^2\,x}$$

$$=\dfrac {\sin\,x}{\cos\,x}×\dfrac {1}{\cos\,x}=\tan\,x\,\sec\,x$$

• Consider an example to understand this.

Let $$y=\dfrac{\sqrt x+\sec x}{2}$$

$$\dfrac{dy}{dx}=\dfrac{d}{dx}\left[\dfrac{\sqrt x+\sec x}{2}\right]$$

$$=\dfrac{1}{2}\dfrac{d}{dx}\left[\sqrt x+\sec x\right]$$

$$=\dfrac{1}{2}\left[\dfrac{d}{dx}(\sqrt x)+\dfrac{d}{dx}(\sec x)\right]$$

$$=\dfrac{1}{2}\left[\dfrac{1}{2 \sqrt x}+\sec x\,\tan x\right]$$

#### If $$y = \dfrac {2+sec\,x}{\sqrt x}$$, find  $$\dfrac {dy}{dx}$$

A $$\dfrac {x\,sec\,x\;tan\,x+cos^2\,x+2}{x}$$

B $$\dfrac {x\,sec^2\,x\;+tan\,x}{\sqrt x}$$

C $$\dfrac {2\,x\,sec\,x\;tan\,x-2-sec\,x}{2\,x\;\sqrt x}$$

D $$\dfrac {2\,x\,sec\,x+x^2\;tan\,x}{\,x^2}$$

×

$$y = \dfrac {2+sec\,x}{\sqrt x}$$

$$\Rightarrow\dfrac {dy}{dx}=\dfrac {d}{dx}\left (\dfrac {2+sec\,x}{\sqrt x}\right)$$

$$= \dfrac {\sqrt x\dfrac {d}{dx}(2+sec\,x)-(2+sec\,x)\dfrac {d}{dx}\,\sqrt x}{x}\rightarrow$$Quotient Rule

$$= \dfrac {\sqrt x× sec\,x\,tan\,x-\dfrac {2+sec\,x}{2\sqrt x}\,}{x}$$

$$= \dfrac {2x\; sec\,x\,tan\,x-2-sec\,x}{2x\sqrt x}$$

### If $$y = \dfrac {2+sec\,x}{\sqrt x}$$, find  $$\dfrac {dy}{dx}$$

A

$$\dfrac {x\,sec\,x\;tan\,x+cos^2\,x+2}{x}$$

.

B

$$\dfrac {x\,sec^2\,x\;+tan\,x}{\sqrt x}$$

C

$$\dfrac {2\,x\,sec\,x\;tan\,x-2-sec\,x}{2\,x\;\sqrt x}$$

D

$$\dfrac {2\,x\,sec\,x+x^2\;tan\,x}{\,x^2}$$

Option C is Correct

# Derivative of cosecant Function(cosec x)

$$\dfrac {d}{dx}(cosec\,x)=-cosec\,x\;cot\,x$$     (where $$x$$ is in radian)

The derivative of $$cosec\,x$$ with respect to $$x\; is-cosec\,x\,cot\,x$$

Proof:

$$\dfrac {d}{dx}(cosec\,x)=\dfrac {d}{dx}\left (\dfrac {1}{sin\,x}\right)$$

$$=\dfrac {sin\,x\dfrac {d}{dx}1-1\dfrac {d}{dx}(sin\,x)}{sin^2\,x}$$

$$=\dfrac {0-cos\,x}{sin^2\,x}=\dfrac {-cos\,x}{sin\,x}×\dfrac {1}{sin\,x}$$

$$=-cot\,x\;cosec\,x$$

• Consider an example to understand this.

Let $$y=\dfrac{x^2+cosec \,x}{x}$$

$$\dfrac{dy}{dx}=\dfrac{d}{dx}\left[\dfrac{x^2+cosec \,x}{x}\right]$$

$$=\dfrac{d}{dx}\left[\dfrac{x^2}{x}+\dfrac{cosec \,x}{x}\right]$$

$$=\dfrac{d}{dx}\left[x+\dfrac{cosec \,x}{x}\right]$$

$$=\dfrac{d}{dx}(x)+\dfrac{d}{dx}\left[\dfrac{cosec \,x}{x}\right]$$

$$=1+\dfrac{x\frac{d}{dx}(cosec\,x)-cosec\,x\frac{d}{dx}(x)}{x^2}$$

$$=1+\dfrac{x(-cosec\,x\;cot\,x)-cosec\,x(1)}{x^2}$$

$$=1-\dfrac{(x\,cosec\,x\;cot\,x+cosec\,x)}{x^2}$$

$$=1-\left[\dfrac{cosec\,x\;[x\,cot\,x+1]}{x^2}\right]$$

#### If $$y=\dfrac {3\,cosec\,x}{x}$$, find  $$\dfrac {dy}{dx}$$

A $$\dfrac {-3\,x\;cosec\,x\;cot\,x-3cosec\,x}{x^2}$$

B $$\dfrac {2\,x\;cosec\,x\;cot\,x+x\,sin\,x}{x^2}$$

C $$\dfrac {x\;sin\,x+cos^2x}{x\,cosec\,x}$$

D $$\dfrac {x\;cos^2\,x+sin^2x}{2x\,cosec\,x}$$

×

$$y=\dfrac {3\,cosec\,x}{x}$$

$$\Rightarrow \dfrac {dy}{dx}= \dfrac {d}{dx}\left ( \dfrac {3\,cosec\,x}{x}\right)$$

$$=3 \dfrac {d}{dx}\left ( \dfrac {cosec\,x}{x}\right)$$

$$=3 \left [ \dfrac {x\,\dfrac {d}{dx}(cosec\,x)-cosec\,x\dfrac {dx}{dx}} {x^2} \right]$$

$$=3 \left [ \dfrac {-x\,cosec\,x\;cot\,x-cosec\,x} {x^2} \right]$$

$$=\dfrac {-3\,x\;cosec\,x\;cot\,x-3cosec\,x}{x^2}$$

### If $$y=\dfrac {3\,cosec\,x}{x}$$, find  $$\dfrac {dy}{dx}$$

A

$$\dfrac {-3\,x\;cosec\,x\;cot\,x-3cosec\,x}{x^2}$$

.

B

$$\dfrac {2\,x\;cosec\,x\;cot\,x+x\,sin\,x}{x^2}$$

C

$$\dfrac {x\;sin\,x+cos^2x}{x\,cosec\,x}$$

D

$$\dfrac {x\;cos^2\,x+sin^2x}{2x\,cosec\,x}$$

Option A is Correct

# Derivative of Tangent Function(tan x)

$$\dfrac{d}{dx} (\tan\; x) = \sec^2 x$$  where $$x$$ is in radians.

The derivative of $$\tan\;x$$ with respect to $$x$$  is $$\sec^2 \;x$$.

Prooof:

$$\dfrac{d}{dx} (\tan\; x) = \dfrac{d}{dx}\; \left(\dfrac{\sin \; x}{\cos \; x}\right) =\underbrace{ \dfrac{\cos \; x \dfrac{d}{dx} \sin \; x - \sin \; x \dfrac{d}{dx} (\cos \; x)}{\cos ^2 \; x}}_\text{Quotient rule }$$

$$=\;\; \dfrac{ \;cos^2x\; + sin ^2\;x}{cos^2\;x} = \dfrac{1}{cos ^2\; x} = sec^2\; x$$.

$$\therefore\; \dfrac{d}{dx} (tan\; x) = sec^2\; x$$.

• Consider an example to understand this.

Let $$y=\dfrac{1}{\sqrt x\tan x}$$

$$\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{1}{\sqrt x\tan x}\right)$$

$$=\dfrac{(\sqrt x\tan x)\frac{d}{dx}(1)-(1)\frac{d}{dx}(\sqrt x\tan x)}{(\sqrt x\tan x)^2}$$   [Quotient rule]

$$=\dfrac{\sqrt x\tan x(0)-\left[\tan x\frac{d}{dx}(\sqrt x)+\sqrt x\frac{d}{dx}(\tan x)\right]}{x\tan^2x}$$

$$=\dfrac{-\tan x\left(\frac{1}{2\sqrt x}\right)-\sqrt x\sec^2x}{x\tan^2x}$$

$$=\dfrac{\sqrt x\left[-\frac{\tan x}{2x}-\sec^2x\right]}{x\tan^2 x}$$

$$=\dfrac{-\left(\frac{\tan x}{2x}+\sec^2x\right)}{\sqrt x\tan^2x}$$

#### If $$y = 2 \sqrt x\; tan \; x$$ then find $$\dfrac{dy}{dx}$$.

A $$\dfrac{2 x \; sec ^2x\; + \; tan \;x}{\sqrt x}$$

B $$\dfrac{2 x\;tan \; x + x^2}{2 x}$$

C $$\dfrac{x\; cos \; x - sin ^2x}{x^2}$$

D $$\dfrac{2 x \; sec^2 x + tan^2 x}{x^2}$$

×

$$y = 2 \sqrt x\; tan \; x$$

$$\Rightarrow\;\;\dfrac{dy}{dx} = \dfrac{d}{dx} \left(2 \sqrt x\; tan \; x\right)$$

$$= \;\; 2 \dfrac{d}{dx} (\sqrt x \; tan \; x) =\underbrace {2\left[\sqrt x \dfrac{d}{dx}(tan\; x) + tan\; x \dfrac{d}{dx} (\sqrt x\;)\right] }_\text{Product rule }$$

$$=2 \left[\sqrt x\; sec^2 x + tan \;x × \dfrac{1}{2 \sqrt x} \right]$$

$$= \; 2 \sqrt x \; sec^2x + \dfrac{tan \; x}{\sqrt x}\; =\;\dfrac{2x\; sec^2 x + tan \; x}{\sqrt x}$$

### If $$y = 2 \sqrt x\; tan \; x$$ then find $$\dfrac{dy}{dx}$$.

A

$$\dfrac{2 x \; sec ^2x\; + \; tan \;x}{\sqrt x}$$

.

B

$$\dfrac{2 x\;tan \; x + x^2}{2 x}$$

C

$$\dfrac{x\; cos \; x - sin ^2x}{x^2}$$

D

$$\dfrac{2 x \; sec^2 x + tan^2 x}{x^2}$$

Option A is Correct

# Derivatives of Cotangent Function(cot x)

•  $$\dfrac{d}{dx}(\cot \; x) = -cosec^2 x$$     ( where $$x$$ is in radians)
• The derivative of $$\cot \; x$$ with respect to $$x$$ is $$-cosec^2x$$

Proof:

$$\dfrac{d}{dx} (\cot \; x) = \dfrac{d}{dx}\left(\dfrac{\cos\; x}{\sin \; x}\right)$$

$$=\;\dfrac{\sin \; x × \dfrac{d}{dx}(\cos\; x) - \cos \; x \dfrac{d}{dx}{(\sin \; x)}}{\sin ^2x}$$

$$=\dfrac{-\sin^2\; x \;-\cos ^2 \; x }{\sin ^2\; x}$$

$$= -1/\sin^2\;x$$

$$= - cosec^2 x$$

$$\therefore\; \dfrac{d}{dx}(\cot \; x ) = - cosec^2 \; x$$

• Consider an example to understand this.

Let $$y=2x+3x\,\cot x$$

$$\dfrac{dy}{dx}=\dfrac{d}{dx}(2x+3x\,\cot x)$$

$$=\dfrac{d}{dx}(2x)+\dfrac{d}{dx}(3x\,\cot x)$$

$$=2+3x\dfrac{d}{dx}(\cot x)+\cot x\dfrac{d}{dx}(3x)$$

$$=2+3x(-cosec^2x)+\cot x(3)$$

$$=2+3\cot x-(3x)\,cosec^2x$$

#### If  $$y = \dfrac{x}{ 2 + cot \;x}$$ find $$\dfrac{dy}{dx}$$.

A $$\dfrac{2 + cot \;x + x \; cosec^2\;x}{(2 + cot \; x)^2}$$

B $$\dfrac{2x\; cot \; x + cosec^2 \; x }{ 2 + cot \; x}$$

C $$\dfrac{2 x\;cosec^2x-sin\; x}{cos \; x}$$

D $$\dfrac{x^3 cos ^2 x +sin \; x }{sin ^2x}$$

×

$$y = \dfrac{x}{2 + cot \; x }\;$$

$$\Rightarrow \dfrac{dy}{dx}= \dfrac{d}{dx} \left(\dfrac{x}{2 + cot \; x }\right)$$

Using Quotient rule:

$$=\; \dfrac{(2 + cot\; x) \dfrac{dx}{dx}-x \dfrac{d}{dx} (2 + cot \; x) }{(2+cot \; x)^2}$$

$$=\;\;\dfrac{(2+cot \; x) × 1 - x × (-cosec^2x)}{(2+cot \; x)^2}$$

$$=\;\; \dfrac{2 + cot \;x + x \; cos ec^2 x }{(2+cot\; x)^2}$$

### If  $$y = \dfrac{x}{ 2 + cot \;x}$$ find $$\dfrac{dy}{dx}$$.

A

$$\dfrac{2 + cot \;x + x \; cosec^2\;x}{(2 + cot \; x)^2}$$

.

B

$$\dfrac{2x\; cot \; x + cosec^2 \; x }{ 2 + cot \; x}$$

C

$$\dfrac{2 x\;cosec^2x-sin\; x}{cos \; x}$$

D

$$\dfrac{x^3 cos ^2 x +sin \; x }{sin ^2x}$$

Option A is Correct

# Finding Derivative of Composite Function using Chain Rule

Composite function

• A composite function is a function of another function.

$$\underbrace{t(x)}_{\text{Composite function}}=\underbrace{f(g(x))}_{\text{Function f of function t}}$$

• Its derivative can be found out by finding the derivative of function $$f$$ multiplied with the derivative of function $$g.$$

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  $$y = f(u)$$ 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 $$f(x) = \cos\,x$$

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

If $$h(x) = \cos (\sin\,x)$$

then $$h'(x) = -\sin (\sin\,x)× \cos\,x$$

$$= - \sin (\sin\,x) \,\cos\,x$$

#### 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

#### If $$y= sin (sin\,x^2)$$ find $$y'$$ .

A $$2x\,cos \,x^2\,cos(sin\,x^2)$$

B $$0$$

C $$1$$

D $$sin\,x^2$$

×

$$=\dfrac{d}{dx} sin(sin\,x^2)$$

$$\Rightarrow cos (sin\,x^2) × \dfrac{d}{dx} (sin\,x^2)$$

$$\Rightarrow cos (sin\,x^2) × cos \,x^2 ×\dfrac{d}{dx} x^2$$

$$\Rightarrow cos (sin\,x^2) × cos\,x^2 ×2\,x$$

$$\Rightarrow2x\,cos \,x^2\,cos(sin\,x^2)$$

### If $$y= sin (sin\,x^2)$$ find $$y'$$ .

A

$$2x\,cos \,x^2\,cos(sin\,x^2)$$

.

B

$$0$$

C

$$1$$

D

$$sin\,x^2$$

Option A is Correct