Informative line

### Application Of Derivative Of Trigonometric Functions

Learn equation of tangent line to a curve involving trigonometric function at a point on it. Practice horizontal tangent for trigonometric curves and finding higher order derivatives of sin x by observing a Pattern.

# Equation of Tangent Line to a Curve Involving Trigonometric Function

The equation of tangent to the curve $$y=f(x)$$ at any point $$(a,f(a))$$ on it, is given by

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

where $$f'(a)$$ is the derivative of function $$f$$ at $$x=a$$ and represents then slope of the tangent.

For example, consider $$y=sin\,x$$  The equation of tangent at $$\left ( \dfrac {\pi}{4},\dfrac {1}{\sqrt 2} \right)$$ will be

$$y-\dfrac {1}{\sqrt 2}=cos\dfrac {\pi}{4} \left ( x-\dfrac {\pi}{4} \right)$$

$$y-\dfrac {1}{\sqrt 2}=\dfrac {1}{\sqrt 2} \left ( x-\dfrac {\pi}{4} \right)$$

$$\Rightarrow \sqrt 2\,y-1=x-\dfrac {\pi}{4}$$

$$\Rightarrow x-\sqrt 2 y+1-\dfrac {\pi}{4}=0$$

#### Find the equation of tangent line to the curve $$y=(2+x^2)\;(cos\,x)$$ at the point  $$(0, 2)$$  on it.

A $$y=2$$

B $$3x+2y+7=0$$

C $$2x+y-1=0$$

D $$x=18$$

×

$$y=(2+x^2)\,cos\,x$$

$$\Rightarrow \dfrac {dy}{dx}=(2+x^2)\,\dfrac {d}{dx}\,(cos\,x)+(cos\,x)\dfrac {d}{dx}(2+x^2)$$

$$=(2+x^2)×(-sin\,x)+cos\,x×2x$$

$$\dfrac {dy}{dx}\Bigg |_{x=0}=(2+0)×(-sin\,0)+cos\,0×2×0=0$$

Equation of tangent is

$$y-y_1=m(x-x_1)$$

$$\Rightarrow\,y-2=0(x-0)$$

$$\Rightarrow y=2$$

### Find the equation of tangent line to the curve $$y=(2+x^2)\;(cos\,x)$$ at the point  $$(0, 2)$$  on it.

A

$$y=2$$

.

B

$$3x+2y+7=0$$

C

$$2x+y-1=0$$

D

$$x=18$$

Option A is Correct

# Horizontal Tangent for Trigonometric Curves

To find the point at which tangent is horizontal we solve the equation

$$\dfrac {dy}{dx}=0$$  i.e. Slope = 0

(Horizontal lines have slope 0 )  #### Find the point at which tangent to the curve  $$y=(2\,sinx-x) \left ( \dfrac {-\pi}{2}\leq x\leq \dfrac {\pi}{2} \right)$$ is horizontal.

A $$\left ( \dfrac {\pi}{3},\;\sqrt 3 - \dfrac {\pi}{3} \right)$$

B $$\left ( \dfrac {\pi}{6},\;\pi-\sqrt 3 \right)$$

C $$(1,-2)$$

D $$(\pi,6)$$

×

For horizontal tangent, $$\dfrac {dy}{dx}=0$$

$$\Rightarrow \dfrac {d}{dx}(2\,sin\,x-x)=0$$

$$\Rightarrow 2\,cos\,x-1=0$$

$$\Rightarrow cos\,x=\dfrac {1}{2}$$

$$\Rightarrow x=\dfrac {\pi}{3}$$

$$\therefore$$ Required point     $$\left (\dfrac {\pi}{3},\;\underbrace {\sqrt 3 - \dfrac {\pi}{3}}_{y(\pi/3)} \right)$$

### Find the point at which tangent to the curve  $$y=(2\,sinx-x) \left ( \dfrac {-\pi}{2}\leq x\leq \dfrac {\pi}{2} \right)$$ is horizontal.

A

$$\left ( \dfrac {\pi}{3},\;\sqrt 3 - \dfrac {\pi}{3} \right)$$

.

B

$$\left ( \dfrac {\pi}{6},\;\pi-\sqrt 3 \right)$$

C

$$(1,-2)$$

D

$$(\pi,6)$$

Option A is Correct

# Nth Derivative of sin(mx)

$$\Rightarrow f(x) =sin\,m\,x$$

$$\Rightarrow f'(x) = m\,cos\,m\,x$$

$$\Rightarrow m\,sin \left(m\,x+\dfrac{\pi}{2}\right)$$

$$f''(x) = -m^2\,sin\,m\,x \Rightarrow m^2 \,sin(m\,x+\pi)$$

$$f''' (x) = -m^3\,cos\,m\,x\Rightarrow m^3\,sin \left(\dfrac{3\pi}{2}+m\,x\right)$$

Nth derivative $$sin\,m\,x = m^n\,sin \left(m\,x +\dfrac{n\,x}{2}\right)$$

#### Find 7 th   derivative of  $$sin\,2\,x$$ .

A $$2^6\,cos\,2x$$

B $$-2^7 \,cos\,2x$$

C $$cos\,2x$$

D $$0$$

×

Here m=2 ,       n= 7

$$\Rightarrow\dfrac{d^7y}{d\,x^7} = 2^7\,sin\,\left(2x+\dfrac{7\,\pi}{2}\right)$$

$$= -2^7\,cos\,2\,x$$

### Find 7 th   derivative of  $$sin\,2\,x$$ .

A

$$2^6\,cos\,2x$$

.

B

$$-2^7 \,cos\,2x$$

C

$$cos\,2x$$

D

$$0$$

Option B is Correct

# Nth Derivative of cos(mx)

$$\because f(x) =cos\,m\,x$$

$$\therefore f'(x) = -m\,sin\,m\,x$$

$$= m\,cos\left(m\,x-\dfrac{\pi}{2}\right)$$

$$f''(x) = -m^2\,cos (m\,x)$$

$$= m^2cos(m\,x-\pi)$$

$$f''' (x) = m^3\,sin(m\,x)$$

$$= m^3 \,cos\left(m\,x-\dfrac{3\,\pi}{2}\right)$$

$$N^{th}$$ derivative of  $$cos\,m\,x$$       (so $$\forall\,n\in I$$)

$$=m^n\,cos\left(m\,x-\dfrac{n\,\pi}{2}\right)$$

#### Find 7th derivative of $$cos\,3\,x$$ .

A $$0$$

B $$3^7\,cos\,3\,x$$

C $$3^7\,sin\,\,x$$

D $$3^7\,sin\,3\,x$$

×

Here m= 3,    n= 7

$$\Rightarrow\dfrac{d^7\,y}{d\,x^7} = 3^7 \,cos\left(3x-\dfrac{7\,\pi}{2}\right)$$

$$= 3^7\,sin\,3\,x$$

### Find 7th derivative of $$cos\,3\,x$$ .

A

$$0$$

.

B

$$3^7\,cos\,3\,x$$

C

$$3^7\,sin\,\,x$$

D

$$3^7\,sin\,3\,x$$

Option D is Correct

# Finding Higher Order Derivatives of sin x by observing a Pattern

Observe that:

$$\dfrac {d}{dx}(sin\,x)=cos\,x\Rightarrow\dfrac {d^2}{dx^2}(sinx)=-sin\,x\Rightarrow\dfrac {d^3y}{dx^3}=-cos\,x$$

$$\Rightarrow\dfrac {d^4y}{dx^4}=sin\,x\rightarrow$$ This pattern will be repeated.

If $$y=sin\,x$$

$$\therefore$$  $$\dfrac {d^ny}{dx^n}=sin\,x$$ if $$n$$ is a multiple of $$4$$

$$\dfrac {d^ny}{dx^n}=cos\,x$$ if $$n$$ is a multiple of $$4n+1$$

$$\dfrac {d^ny}{dx^n}=-sin\,x$$ if $$n$$ is a multiple of $$4n+2$$

$$\dfrac {d^ny}{dx^n}=-cos\,x$$ if $$n$$ is a multiple of $$4n+3$$

#### Find the $$35^{th}$$ derivative of $$y=sin\,x$$

A $$sin\,x$$

B $$-cos\,x$$

C $$tan\,x$$

D $$sec^2\,x$$

×

$$35=4×8+3\rightarrow$$ is of the form $$4n+3$$

$$\therefore \dfrac {d^{35}}{dx^{35}}\,(sin\,x)=-cos\,x$$

### Find the $$35^{th}$$ derivative of $$y=sin\,x$$

A

$$sin\,x$$

.

B

$$-cos\,x$$

C

$$tan\,x$$

D

$$sec^2\,x$$

Option B is Correct

# Finding Higher Order Derivative of cos x by Observing a Pattern

Observe that:

$$\dfrac {d}{dx}(cos\,x)=-sin\,x$$

$$\Rightarrow\dfrac {d^2}{dx^2}(cos\,x)=-cos\,x$$

$$\Rightarrow\dfrac {d^3}{dx^3}\,(cos\,x)=sin\,x$$

$$\Rightarrow\dfrac {d^4}{dx^4}\,(cos\,x)=cos\,x$$

If $$y=cos\,x$$

$$\therefore$$  $$\dfrac {d^ny}{dx^n}=cos\,x$$ if $$n$$ is a multiple of $$4$$

$$\dfrac {d^ny}{dx^n}=-sin\,x$$ if $$n$$ is a multiple of $$4n+1$$

$$\dfrac {d^ny}{dx^n}=-cos\,x$$ if $$n$$ is a multiple of $$4n+2$$

$$\dfrac {d^ny}{dx^n}=sin\,x$$ if $$n$$ is a multiple of $$4n+3$$

#### Find the $$81^{th}$$ derivative of $$y=cos\,x$$.

A $$-sin\,x$$

B $$cos\,x$$

C $$tan\,x$$

D $$sec\,x$$

×

$$81=4×20+1\rightarrow$$ is of the form $$4n+1$$

$$\therefore \dfrac {d^{81}}{dx^{81}}\,(cos\,x)=-sin\,x$$

### Find the $$81^{th}$$ derivative of $$y=cos\,x$$.

A

$$-sin\,x$$

.

B

$$cos\,x$$

C

$$tan\,x$$

D

$$sec\,x$$

Option A is Correct