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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-

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

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

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

Illustration Questions

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]

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

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

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

Illustration Questions

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 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]

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

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

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

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

Illustration Questions

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 \(y=2x+3x\,\cot x\)

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

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

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

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

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

 

Illustration Questions

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

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 \(y=\dfrac{\sqrt x+\sec x}{2}\)

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

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

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

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

Illustration Questions

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 \(y=\dfrac{x^2+cosec \,x}{x}\)

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

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

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

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

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

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

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

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

Illustration Questions

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

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

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

Illustration Questions

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

Illustration Questions

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

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