Find the equation of the tangent line and Normal to the curve, the derivative of the function at the given point. Practice to find the point on the curve at which tangent is horizontal (parallel to x axis).

When we are asked to find \(f'(a)\), we first find the derivative of \(f\) with respect to \(x\) and then put \(x=a\). It will not mean the derivative of the value \(f(a)\) because this value will always be 0, since \(f(a)\) is a constant.

**Note: **

Do not simplify the expression of derivative, just put the value of \(x\) as soon as differentiation step is complete.

A \(\dfrac{25}{18}\)

B \(\dfrac{19}{16}\)

C \(\dfrac{4}{3}\)

D \(\dfrac{-7}{8}\)

Let \(y=f(x)\) be a curve and \((a,\,f(a))\) be any point on it. The equation of tangent at this point is given by.

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

A \(5x+y+17=0\)

B \(x+2y+18=0\)

C \(2x-y-3=0\)

D \(4x+11y-7=0\)

- Given the slope, equate the derivative of the function to that slope and solve for \(x\), hence find the point.

A \((0,\,-1)\,\;or\;\;(2,\,3)\)

B \((0,\,1)\,\;or\;\;(3,\,5)\)

C \(\left(5,\,\dfrac{3}{2}\right)\,\;or\;\;(1,\,0)\)

D \((2,\,3)\,\;or\;\;(-1,\,0)\)

If we are given a mixed function and its derivative is desired at particular values of \(x\), then use the derivative of component function given at that \(x\) and appropriate rule of differentiation.

A \(\dfrac{-43}{4}\)

B \(\dfrac{76}{3}\)

C \(\dfrac{18}{5}\)

D –19

If two lines are perpendicular to each other then the product of their slopes is –1

i.e. \(m_1m_2=-1\) if \(m_1,\,m_2\) are slopes of \(\bot\) lines.

Let \(y=f(x)\) be a curve and \((a,\,f(a))\) be any point on it, then the equation of normal at this point is

\(y-f(a)={\dfrac{-1}{f'(a)}}(x-a)\)

A \(x+7y-84=0\)

B \(3x+8y-1=0\)

C \(2x+5y-48=0\)

D \(11x+2y+7=0\)

To find the point where tangent is horizontal or parallel to \(x\) axis.

- Take the derivative, find \(\dfrac{dy}{dx}\)
- Put \(\dfrac{dy}{dx}=0\)
- Solve for \(x\).

A \(x=2\;or\;x=3\)

B \(x=8\;or\;x=1\)

C \(x=-7\;or\;x=5\)

D \(x=17\;or\;x=-18\)

Sometimes equation of a tangent is desired whose contact point is not given but its slope is given, either directly or indirectly.

- First we find the contact point, and then find equation of tangent.

A \(4x-y-4=0 \;\;\&\;\;4x-y=0\)

B \(2x+3y+4=0 \;\;\&\;\;x-y=0\)

C \(4x+y+7=0 \;\;\&\;\;x+2y+8=0\)

D \(x+3y=0 \;\;\&\;\;2x+y+7=0\)

If slope of tangent at some point of a graph is given, equate it to the derivative at that point. If the function contains some parameter, they can be solved.

A \(y=2x^2+5x+3\)

B \(y=5x^2+8x+7\)

C \(y=7x^2=x-9\)

D \(y=3x^2+8x+18\)