Informative line

### Application Of Derivative

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).

# Finding the Derivatives at Particular Values of x

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.

#### If $$f(t)=\dfrac{t^2-5t-1}{t^3}$$, find the value of $$f'(2)$$.

A $$\dfrac{25}{18}$$

B $$\dfrac{19}{16}$$

C $$\dfrac{4}{3}$$

D $$\dfrac{-7}{8}$$

×

$$f(t)=\dfrac{t^2-5t-1}{t^3}$$

$$\Rightarrow$$ $$f'(t)$$ $$=\underbrace{\dfrac{t^3\dfrac{d}{dt}(t^2-5t-1)-(t^2-5t-1)\dfrac{d}{dt}t^3}{(t^3)^2}}_{Use \,Quotient\,Rule}$$

$$=\dfrac{t^3×(2t-5)-(t^2-5t-1)×3t^2}{t^6}$$

$$f'(2)$$$$\dfrac{8×(4-5)-(4-10-1)×3×4}{2^6}$$

$$=\dfrac{-8+84}{64}$$

$$=\dfrac{76}{64}$$

$$=\dfrac{19}{16}$$

### If $$f(t)=\dfrac{t^2-5t-1}{t^3}$$, find the value of $$f'(2)$$.

A

$$\dfrac{25}{18}$$

.

B

$$\dfrac{19}{16}$$

C

$$\dfrac{4}{3}$$

D

$$\dfrac{-7}{8}$$

Option B is Correct

# Equation of Tangent to a Curve at a given Point

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.

$$y-f(a)=$$ $$f'(a)$$ $$(x-a)$$

#### Find the equation of tangent to the curve $$y=\dfrac{1+2x}{3-4x}$$ at the point $$(1,\,-1)$$ on it.

A $$5x+y+17=0$$

B $$x+2y+18=0$$

C $$2x-y-3=0$$

D $$4x+11y-7=0$$

×

$$y=\dfrac{1+2x}{3-4x}$$

$$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{(3-4x)×2-(1+2x)×-4}{(3-4x)^2}$$

$$=\dfrac{6-8x-4+8x}{(3-4x)^2}$$

$$=\dfrac{2}{(3-4x)^2}$$

$$\Rightarrow$$  $$y'$$ at $$(1,\,-1)$$ $$=\dfrac{2}{(3-4)^2}=2$$

Equation of tangent is  $$y-(-1)=$$ $$y'$$$$(x-1)$$

$$\Rightarrow\,y+1=2(x-1)$$

$$\Rightarrow\,2x-y-3=0$$

### Find the equation of tangent to the curve $$y=\dfrac{1+2x}{3-4x}$$ at the point $$(1,\,-1)$$ on it.

A

$$5x+y+17=0$$

.

B

$$x+2y+18=0$$

C

$$2x-y-3=0$$

D

$$4x+11y-7=0$$

Option C is Correct

# Finding the Point on the Curve when Slope of Tangent is given

• Given the slope, equate the derivative of the function to that slope and solve for $$x$$, hence find the point.

#### Find the point(s) on the curve $$y=\dfrac{x+1}{x-1}$$ at which the slope of tangent is –2.

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

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

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

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

×

$$y=\dfrac{x+1}{x-1}$$

$$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{(x-1)×1-(x+1)×1}{(x-1)^2}$$

$$\therefore\,\dfrac{dy}{dx}=\dfrac{-2}{(x-1)^2}$$

Now

$$\dfrac{dy}{dx}=-2$$

$$\Rightarrow\,\dfrac{-2}{(x-1)^2}=-2$$

$$\Rightarrow\,(x-1)^2=1$$

$$\Rightarrow\,x-1=\pm1$$

$$\Rightarrow\,x=2\;or\;x=0$$

$$\therefore\,$$ required points are $$(0,\,-1)\;or\;(2,\,3)$$

### Find the point(s) on the curve $$y=\dfrac{x+1}{x-1}$$ at which the slope of tangent is –2.

A

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

.

B

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

C

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

D

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

Option A is Correct

# Evaluating Derivatives of Mixed Function at Particular Values of x

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.

#### If $$f$$ and $$g$$ are differentiable functions such that $$f'(3)$$$$=4$$, $$f(3)=5$$, $$g'(3)$$$$=7$$, $$g(3)=-2$$ then the value of $$\left(\dfrac{f}{g}\right)' at\, x=3\, is$$

A $$\dfrac{-43}{4}$$

B $$\dfrac{76}{3}$$

C $$\dfrac{18}{5}$$

D –19

×

$$\left(\dfrac{f}{g}\right)'=\dfrac{gf'-fg'}{g^2}$$

(Quotient Rule)

$$\left(\dfrac{f}{g}\right)'\Bigg|_{x=3}=\dfrac{g(3)f'(3)-f(3)g'(3)}{(g(3))^2}$$

$$=\dfrac{-2×4-5×7}{(-2)^2}$$

$$=\dfrac{-8-35}{4}$$

$$=\dfrac{-43}{4}$$

### If $$f$$ and $$g$$ are differentiable functions such that $$f'(3)$$$$=4$$, $$f(3)=5$$, $$g'(3)$$$$=7$$, $$g(3)=-2$$ then the value of $$\left(\dfrac{f}{g}\right)' at\, x=3\, is$$

A

$$\dfrac{-43}{4}$$

.

B

$$\dfrac{76}{3}$$

C

$$\dfrac{18}{5}$$

D

–19

Option A is Correct

# Equation of Normal to a Curve at a given Point

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

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

#### Find the equation of normal to the curve $$y=3x-2\sqrt x$$ at the point $$(4,\,8)$$ on it.

A $$x+7y-84=0$$

B $$3x+8y-1=0$$

C $$2x+5y-48=0$$

D $$11x+2y+7=0$$

×

$$y=3x-2\sqrt x$$

$$\Rightarrow$$ $$y'$$ $$=3-\dfrac{2}{2\sqrt x}$$

$$\Rightarrow$$$$y'$$$$=3-\dfrac{1}{\sqrt x}$$

$$\Rightarrow$$ $$y'$$ at $$(4,\,8)=3-\dfrac{1}{2}=\dfrac{5}{2}$$

Equation of normal is $$y-8=\dfrac{-1}{y'}(x-4)$$

$$\Rightarrow\,y-8=\dfrac{-1}{\dfrac{5}{2}}(x-4)$$

$$\Rightarrow\,y-8=\dfrac{-2}{5}(x-4)$$

$$\Rightarrow\,5y-40=-2x+8$$

$$\Rightarrow\,2x+5y-48=0$$

### Find the equation of normal to the curve $$y=3x-2\sqrt x$$ at the point $$(4,\,8)$$ on it.

A

$$x+7y-84=0$$

.

B

$$3x+8y-1=0$$

C

$$2x+5y-48=0$$

D

$$11x+2y+7=0$$

Option C is Correct

# Finding the Point on the Curve at which Tangent is Horizontal(Parallel to x axis)

To find the point where tangent is horizontal or parallel to $$x$$ axis.

1. Take the derivative, find $$\dfrac{dy}{dx}$$
2. Put $$\dfrac{dy}{dx}=0$$
3. Solve for $$x$$.

#### For what value of $$x$$ will the graph of  $$y=2x^3-15x^2+36x+1$$ have a horizontal tangent ?

A $$x=2\;or\;x=3$$

B $$x=8\;or\;x=1$$

C $$x=-7\;or\;x=5$$

D $$x=17\;or\;x=-18$$

×

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

$$y=2x^3-15x^2+36x+1$$

$$\Rightarrow\,\dfrac{dy}{dx}=6x^2-30x+36$$

then  $$\dfrac{dy}{dx}=0$$

$$\Rightarrow\,6x^2-30x+36=0$$

$$\Rightarrow\,x^2-5x+6=0$$

$$\Rightarrow\,(x-2)\,(x-3)=0$$

$$\Rightarrow\,x=2\;or\;x=3$$

### For what value of $$x$$ will the graph of  $$y=2x^3-15x^2+36x+1$$ have a horizontal tangent ?

A

$$x=2\;or\;x=3$$

.

B

$$x=8\;or\;x=1$$

C

$$x=-7\;or\;x=5$$

D

$$x=17\;or\;x=-18$$

Option A is Correct

# Finding the Equation of Tangent to a Curve using the given Slope

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.

#### Find the equation of tangent line to $$y=x^3+x-2$$ which is parallel to straight line $$y=4x-1$$

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 two lines are parallel then there slopes are equal. So the required tangent has slope = 4.

(Same as slope of $$y=4x-1$$)

$$\therefore\,\dfrac{dy}{dx}=4$$

$$\Rightarrow\,3x^2+1=4$$

$$\Rightarrow\,3x^2=3$$

$$\Rightarrow\,x=\pm1$$

$$\therefore$$ The point at which tangent have slope 4 are $$(1,\,0)$$ and $$(-1,\,-4)$$.

$$\therefore$$ There are two tangents -

(1)  $$y-0=4(x-1)$$

$$\Rightarrow\,4x-y-4=0$$

(2)  $$y+4=4(x+1)$$

$$\Rightarrow\,4x-y=0$$

(There can be more then one tangent to a curve of same slope).

### Find the equation of tangent line to $$y=x^3+x-2$$ which is parallel to straight line $$y=4x-1$$

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

$$x+3y=0 \;\;\&\;\;2x+y+7=0$$

Option A is Correct

# Identifying the Parameters of a Curve

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 curve has the equation of the form  $$y=ax^2+bx+c$$. It has slope 9 at $$x=1$$, slope 13 at $$x=2$$ and passes through $$(0,\,3)$$, find the curve.

A $$y=2x^2+5x+3$$

B $$y=5x^2+8x+7$$

C $$y=7x^2=x-9$$

D $$y=3x^2+8x+18$$

×

Slope = 9 at $$x=1$$

$$\Rightarrow\,\dfrac{dy}{dx}\Bigg|_{x=1}=9$$

$$\Rightarrow\,2ax+b\Bigg|_{x=1}=9$$

$$\Rightarrow\,2a+b=9\to$$  (1)

Slope = 13 at $$x=2$$

$$\Rightarrow\,\dfrac{dy}{dx}\Bigg|_{x=2}=13$$

$$\Rightarrow\,4a+b=13\to$$  (2)

Passes through $$(0,\,3)$$

$$\Rightarrow3=c$$

Solving equation (1) and (2) for $$'a'$$ and $$'b'$$  we get

$$a=2,\,\;b=5$$

$$\therefore$$ the required curve is $$y=2x^2+5x+3$$.

### A curve has the equation of the form  $$y=ax^2+bx+c$$. It has slope 9 at $$x=1$$, slope 13 at $$x=2$$ and passes through $$(0,\,3)$$, find the curve.

A

$$y=2x^2+5x+3$$

.

B

$$y=5x^2+8x+7$$

C

$$y=7x^2=x-9$$

D

$$y=3x^2+8x+18$$

Option A is Correct