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.

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

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

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

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

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

A \(y=2\)

B \(3x+2y+7=0\)

C \(2x+y-1=0\)

D \(x=18\)

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 )

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

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

C \((1,-2)\)

D \((\pi,6)\)

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

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

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

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

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

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

A \(2^6\,cos\,2x\)

B \(-2^7 \,cos\,2x \)

C \(cos\,2x\)

D \(0\)

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

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

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

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

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

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

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

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

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

A \(0\)

B \(3^7\,cos\,3\,x\)

C \(3^7\,sin\,\,x\)

D \(3^7\,sin\,3\,x\)

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 \(4n+3\)

A \(sin\,x\)

B \(-cos\,x\)

C \(tan\,x\)

D \(sec^2\,x\)

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 \(4n+3\)

A \(-sin\,x\)

B \(cos\,x\)

C \(tan\,x\)

D \(sec\,x\)