Informative line

### Implicit Derivative And Its Application

Practice implicit differentiation problems & examples, Learn equation derivatives at particular values of ?x? and ?y? by implicit differentiation & higher order derivatives of simple function given in the implicit form, orthogonal curves.

# Implicit Differentiation

Functions are many times defined in a  way that $$y$$ is expressed clearly in terms of  $$x$$ ,  we say  $$y$$  is an explicit function of  $$x$$ .

$$y =f(x) \to$$ explicit function

• Some function are  defined by giving a relation between  $$x$$  and  $$y$$.

e.g  $$x^2+y^3=3xy\to$$ implicit relation

We  say $$y$$ is an implicit function of $$x$$  in this case.

• To find  $$\dfrac{dy}{dx}$$, differentiate both sides with respect to $$x$$, treating $$y$$ as a function of  $$x$$

and solve the resulting equation for $$y'$$ .

• $$\dfrac{dy}{dx}$$ may contain terms of $$y$$ when we differentiate implicit functions.
• $$\dfrac{d}{dx}(y^n) = n\,y^{n-1} \dfrac{dy}{dx} \to$$ we treat  $$y$$  as a function of  $$x$$ just like $$f(x)$$ or $$g(x)$$ .

So   $$\dfrac{d}{dx} (f(x))^n = n(f(x))^{n-1} f'(x)$$ by Chain Rule.

• Similarly  $$\dfrac{d}{dx} (y^n) = n\,y^{n-1} y'$$

#### If $$y^3 - 3y +4x = 0$$ , find $$y'$$ by implicit differentiation .

A $$\dfrac{46}{1-y^2}$$

B $$\dfrac{2}{7(1-y^2)}$$

C $$2y+3$$

D $$\dfrac{4}{3(1-y^2)}$$

×

$$y^3 - 3y +4x = 0 \to$$ Differentiating both sides with respect to   $$x$$

$$\Rightarrow 3\,y^2 \dfrac{dy}{dx} - 3\dfrac{dy}{dx} + 4 =0$$    (we often  use $$y'$$ in place of $$\dfrac{dy}{dx}$$ for convenience)

$$\Rightarrow 3\,y^2\, y'- 3y'+4 = 0$$

$$\Rightarrow y'(3y^2-3) = -4$$

$$\Rightarrow y'= \dfrac{-4}{3y^2-3}$$

$$= \dfrac{4}{3(1-y^2)}$$

### If $$y^3 - 3y +4x = 0$$ , find $$y'$$ by implicit differentiation .

A

$$\dfrac{46}{1-y^2}$$

.

B

$$\dfrac{2}{7(1-y^2)}$$

C

$$2y+3$$

D

$$\dfrac{4}{3(1-y^2)}$$

Option D is Correct

# Implicit Differentiation of Mixed Functions

• Sometimes we need to differentiate functions which are implicit and contain non zero terms on both sides of the equation .
• To get  $$\dfrac{dy}{dx}$$  we differentiate both sides with respect to $$x$$ , treating  $$y$$ as a function of  $$x$$  and  $$y$$.
• In most of these cases  $$\dfrac{dy}{dx}$$ will contain both $$x$$ and  $$y$$ terms.

#### If  $$y\ sin\,x^3 = x\,sin\,y^3$$  then find  $$y'$$  by implicit differentiation.

A $$\dfrac{sin\,x^3 - cos\,x^3}{x^2-y^2}$$

B $$\dfrac{sin\,y^3 + 3y\,cos\,x^3}{2\,y^2+x^3}$$

C $$\dfrac{sin\,y^3 - 3\,x^2y\,cos\,x^3}{sin\,x^3 - 3y^2\,x\,cos\,y^3}$$

D $$\dfrac{cos\,y^3 - 3\,y^2x\,cos\,x^3}{sin\,x^3 +y^3}$$

×

$$y\ sin\,x^3 = x\,sin\,y^3 \to$$ Differentiating  both sides with respect to  $$x$$ .

$$\Rightarrow y(cos\,x^3 × 3x^2) + y'sin\,x^3 = sin\,y^3\,+\,x\,cos\,y^3 × 3\,y^2\,y'$$

$$\Rightarrow 3x^2y\,cos\,x^3 + y'sin\,x^3 = sin\,y^3 + 3y^2\,x \,cos\,y^3\,y'$$

$$\Rightarrow y'[sin\,x^3 - 3y^2 \,x\,cos\,y^3] = sin\,y^3 - 3\,x^2\,y\,cos\,x^3$$

$$\Rightarrow y'= \dfrac{sin\,y^3 - 3x^2y \,cos\,x^3}{sin\,x^3 - 3y^2x \,cos\,y^3}$$

### If  $$y\ sin\,x^3 = x\,sin\,y^3$$  then find  $$y'$$  by implicit differentiation.

A

$$\dfrac{sin\,x^3 - cos\,x^3}{x^2-y^2}$$

.

B

$$\dfrac{sin\,y^3 + 3y\,cos\,x^3}{2\,y^2+x^3}$$

C

$$\dfrac{sin\,y^3 - 3\,x^2y\,cos\,x^3}{sin\,x^3 - 3y^2\,x\,cos\,y^3}$$

D

$$\dfrac{cos\,y^3 - 3\,y^2x\,cos\,x^3}{sin\,x^3 +y^3}$$

Option C is Correct

# Finding the Value of Higher Order Derivatives at Particular Values of x

• If a curve  is given in the implicit form  $$f(x,y)=0$$ , to find the second order derivative at particular  $$x$$   of  $$y$$ with respect to  $$x$$, we first differentiate and find  $$y'$$   at the particular point.
• We differentiate again and obtain  $$y''$$  in terms of  $$y'$$ , $$y$$  and  $$x$$. Now put the values of  $$x,y$$  which are given and  $$y'$$ that is obtained in the previous step.

#### If $$x^2\,y+y^3 =1$$, find the value of $$y''$$  at the point where  $$x=0$$.

A $$\dfrac{-2}{3}$$

B $$-4$$

C $$6$$

D $$8$$

×

When $$x=0$$ , find $$y$$, put  $$x=0$$ in the curve

$$0+y^3 =1$$

$$\Rightarrow y=1$$

$$\therefore \, x=0,\,y=1$$

Differentiate both sides with respect to  $$x$$,

$$\Rightarrow x^2 \,y'+2\,x\,y + 3\,y^2\,y'=0\to(1)$$

Put  $$x=0,\,y=1$$  to find  $$y'$$  at  $$x=0$$

$$\Rightarrow 0+0+3\,y'=0 \,$$

$$\Rightarrow y'=0$$

Differentiate (1) again with respect to   $$x$$ ,

$$\Rightarrow x^2\,y'' + 2\,x\,y'+2\,x\,y'+2\,y+3\,y^2\,y'' + 6\,y\,y'\,y'=0$$

Put  $$x=0 ;\,y=1\,;y'= 0$$

$$\Rightarrow 0+0+2+0+3\,y'' + 0=0$$

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

### If $$x^2\,y+y^3 =1$$, find the value of $$y''$$  at the point where  $$x=0$$.

A

$$\dfrac{-2}{3}$$

.

B

$$-4$$

C

$$6$$

D

$$8$$

Option A is Correct

# Higher Order Derivatives by  Implicit Differentiation

If higher order derivatives of implicit function are desired , we do the following steps:

(1) Differentiate both sides with respect to  $$x$$, treating $$y$$ as a function of  $$x$$.

(2) Again differentiate  both sides with respect to  $$x$$ , taking all terms of  $$y'$$  obtained in (1) as function of  $$x$$ .

(3) Obtain an expression for $$y''$$ , if it contains  $$y'$$  term. Use result of step (1) to express $$y''$$ in terms of $$x$$ and $$y$$ .

#### If  $$2x^2+3y^2=5$$ find $$y''$$ by implicit differentiation .

A   $$\dfrac{-4\,x^2-6\,y^2}{9\,y^3}$$

B $$\dfrac{4\,x^2+6\,y^2}{\,x^3}$$

C   $$\dfrac{7\,x^2+\,y^2}{\,y^3}$$

D   $$\dfrac{2\,x^2-6\,y^2}{5\,y^3}$$

×

$$2\,x^2+3\,y^2=5\to$$ differentiate both sides with respect to  $$x$$.

$$\Rightarrow 4x+6y\,y'=0\;\;----(1)$$again differentiate both sides with respect to  $$x$$.

$$\Rightarrow 4+6y\,y''+6y'\,y'=0$$

$$\Rightarrow 4+6\,y\,y'' +6 \left(\dfrac{-4x}{6\,y}\right)^2 =0$$ [  put  $$y'$$  from (1)  ]

$$\Rightarrow 4+6\,y\,y'' +6 × \dfrac{16\,x^2}{36\,y^2}=0$$

$$\Rightarrow 4+6\,yy'' + \dfrac{8x^2}{3\,y^2}=0\,$$

$$\Rightarrow 12\,y^2 +18\,y^3y'' +8x^2=0$$

$$\Rightarrow 6y^2 +9\,y^3\,y''+4\,x^2=0$$

$$\Rightarrow y'' = \dfrac{-4x^2-6y^2}{9\,y^3}$$

### If  $$2x^2+3y^2=5$$ find $$y''$$ by implicit differentiation .

A

$$\dfrac{-4\,x^2-6\,y^2}{9\,y^3}$$  .

B

$$\dfrac{4\,x^2+6\,y^2}{\,x^3}$$

C

$$\dfrac{7\,x^2+\,y^2}{\,y^3}$$  D

$$\dfrac{2\,x^2-6\,y^2}{5\,y^3}$$  Option A is Correct

# Finding Equation of Tangent with Implicit Differentiation at a Given Point

• Consider an implicit function  $$y$$ of $$x$$ in the form   $$f(x,y)=0$$
• First find  $$y'$$  or  $$\dfrac{dy}{dx}$$  by implicit  function differentiation.
• Evaluate  $$y'$$  at the point $$(x_1,y_1)$$ which is given on the curve.
• Use this value of  $$y'$$  (it may contain both $$x$$and $$y$$)  in the equation of tangent at  $$(x_1,y_1)$$  which is  $$y-y_1=y' (x-x_1)$$

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

A $$4x+7y-1=0$$

B $$4\,x+3\,y-2=0$$

C $$2\,x+3\,y-1=0$$

D $$2\,x+\,y-3=0$$

×

Equation of tangent $$\to y-y_1 = \dfrac{dy}{dx}(x-x_1)$$

where $$(x_1,y_1)$$ is the point $$\to x_1=1, y_1=1$$

Differentiate both sides with respect to  $$x$$.

$$x^2\,y' + 2\,x\,y +y^2+2\,y\,y'\,x +3\,x^2=0$$

Put $$x =1,\,y=1$$

$$\Rightarrow y' +2+1+2\,y'+3=0$$

$$\Rightarrow 3\,y'+6=0$$

$$\Rightarrow y' =-2$$

$$\therefore$$ Equation of tangent $$\to y-1=-2(x-1)$$

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

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

A

$$4x+7y-1=0$$

.

B

$$4\,x+3\,y-2=0$$

C

$$2\,x+3\,y-1=0$$

D

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

Option D is Correct

# Higher Order Derivatives of Simple Function Given in the Implicit Form

•  If  $$x ^2y=2$$ then on differentiating with respect to $$x$$ we get,  $$x^2y' +2\,x\,y=0$$
• Differentiate it again with respect to $$x$$.

$$x^2y''+2\,x\,y' + 2\,x\,y' +2\,y =0$$     $$\Rightarrow x^2\,y'' +4\,xy'+2\,y=0$$

#### If $$x\,y^2=2$$ find $$y''$$.

A $$\dfrac{3y}{4x^2}$$

B $$\dfrac{-(x^3+4x)}{4\,y^2}$$

C $$\dfrac{-(x+x\,y^2)}{\,x}$$

D $$\dfrac{-(y+y^3)}{5\,x}$$

×

$$x\,y^2=2$$

Differentiate both sides with respect to $$x$$

$$\Rightarrow x× 2\,y\,y' +y^2=0\to$$Product Rule

$$\Rightarrow y' = \dfrac{-y^2}{2\,x\,y} = \dfrac{-y}{2x}$$

Differentiate again with respect to  $$x$$

$$\Rightarrow 2\,x\,y\,y'' + 2\,x\,y' ^2 + 2\,y\,y'+2\,y\,y'=0$$

$$\Rightarrow 2\,x\,y\,y'' + 2\,x\,y' ^2 + 4\,y\,y'=0$$

$$\Rightarrow \,x\,y\,y'' + \,x\,y' ^2 + 2\,y\,y'=0$$

$$\Rightarrow y'' = \dfrac{-(x\,y'^2+2\,y\,y')}{xy}$$

$$\Rightarrow y'' = \dfrac{-\left(x\,× \dfrac{y^2}{4\,x^2}+2y× \dfrac{-y}{2x}\right)}{xy}$$

$$\Rightarrow y'' = \dfrac{ \dfrac{y^2}{\,x}- \dfrac{y^2}{4x}}{xy}=\dfrac{4y^2-y^2}{4\,x^2\,y}$$

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

$$\Rightarrow y''=\dfrac{3y}{4x^2}$$

### If $$x\,y^2=2$$ find $$y''$$.

A

$$\dfrac{3y}{4x^2}$$

.

B

$$\dfrac{-(x^3+4x)}{4\,y^2}$$

C

$$\dfrac{-(x+x\,y^2)}{\,x}$$

D

$$\dfrac{-(y+y^3)}{5\,x}$$

Option A is Correct

# Orthogonal Curves

• Two curves are said to be orthogonal if tangent lines at the point of intersection are perpendicular.

$$y= f(x)$$ and $$y= g(x)$$ are orthogonal , tangent are $$\bot$$ at the intersection point P.  • In the figure shown, $$y=f(x)\,\,$$  and $$y=g(x)$$ are orthogonal, tangents are perpendicular at the intersection point P.
• Two family of curves are said to be orthogonal trajectories of  each other if each member of one family is orthogonal to each member of other.
• If two lines with slopes  $$m_1 \,and \,m_2$$  are  $$\bot$$  then  $$m_1\,m_2=-1$$
• Consider two lines  $$\ell_1 \,and\,\ell_2$$   which are  $$\bot$$  to each other.  •  slope of $$\ell_1 = tan\,\theta_1 = m_1$$
•            slope of $$\ell_2 = tan\,\theta_2 = m_2$$

Now $$\theta_2 = 90° +\theta_1$$

$$\Rightarrow tan\,\theta_2 = tan(90°+\theta_1)\,$$

$$\Rightarrow tan\,\theta_2 = -cot\,\theta_1$$

$$\Rightarrow tan\,\theta_2 = \dfrac{-1}{tan\,\theta_1}$$

$$\Rightarrow 1+tan\,\theta_2 \,tan\,\theta_1=0\,$$

$$\Rightarrow 1+m_1m_2=0$$

$$\Rightarrow\, m_1m_2 = -1$$

• Condition of orthogonality  $$\to \dfrac{df}{dx} ×\dfrac{dg}{dx} = -1$$   at the point of intersection .

(for $$\bot$$ lines $$m_1\,m_2=-1$$)

#### Choose the curve which is orthogonal to $$y= 7\,x^2$$.

A $$x^2+2y^2 =1$$

B $$x^2=6y$$

C $$y^2=6\,x$$

D $$3\,y^2+x^2=1$$

×

For the given curve  $$y= 7\,x^2 \to y' =14\,x$$

$$\therefore \dfrac{df}{dx} = 14 \,x$$

Now differentiate the relations given in options  (A), (B), (C), (D).

(A) $$\to 2x+4\,y\,y' =0$$

$$\Rightarrow y' = \dfrac{-x}{2\,y} = \dfrac{d\,g}{d\,x}$$

$$\therefore \dfrac{df}{dx} . \dfrac{dg}{dx} = 14\,x× \dfrac{-x}{2y} = \dfrac{-14\,x^2}{2y} = \dfrac{-14×y}{2× 7\,y}(y=7x^2)$$

$$=-1$$

$$\therefore\;$$ Option (A) is correct.

(B) $$x^2 = 6y$$

$$\Rightarrow 2\,x =6\,y'$$

$$\Rightarrow y' = \dfrac{x}{3} = \dfrac{dg}{dx}$$

$$\therefore \dfrac{df}{dx} . \dfrac{dg}{dx} = 14\,x × \dfrac{x}{3} = \dfrac{14}{3}x^2\neq-1$$

$$\therefore$$ not orthogonal

(C) $$y^2 = 6x$$

$$\Rightarrow 2\,y\,y' =6\,$$

$$\Rightarrow y' = \dfrac{3}{y} = \dfrac{dg}{dx}$$

$$\therefore \dfrac{df}{dx} . \dfrac{dg}{dx} = 14\,x × \dfrac{3}{y} = \dfrac{42\,x}{y}\neq-1$$

$$\therefore$$ not orthogonal

(D) $$3\,y^2+x^2 = 1$$

$$\Rightarrow 6\,y\,y' +2x=0$$

$$\Rightarrow y' = \dfrac{-x}{3\,y} = \dfrac{dg}{dx}$$

$$\therefore \dfrac{df}{dx} . \dfrac{dg}{dx} = 14\,x × \dfrac{(-x)}{3\,y}$$

$$= \dfrac{-14}{3}\dfrac{x^2}{y}$$

$$=\dfrac{-2}{3}\left(x^2=\dfrac{y}{7}\right)\neq-1$$

$$\therefore$$ not orthogonal

### Choose the curve which is orthogonal to $$y= 7\,x^2$$.

A

$$x^2+2y^2 =1$$

.

B

$$x^2=6y$$

C

$$y^2=6\,x$$

D

$$3\,y^2+x^2=1$$

Option A is Correct

# Derivatives at given Values by Implicit Differentiation

• Given any function in the implicit form , to find derivative at particular values of  $$x$$ , we first  differentiate and then put the particular value of  $$x$$  and corresponding  $$y$$ in the expression .
• e.g   If $$(f(x))^3 + (f(x)) = 10 \,x$$  and  $$f(1)=2$$   then differentiate  both sides with respect to  $$x$$ .

$$3(f(x))^2 \,f'(x) + f'(x) =10 \to$$ put  $$x=1$$ .

$$\Rightarrow 3× (f(1))^2 × f'(1)+f' (1) = 10$$

$$\Rightarrow 3× 4× f' (1) + f' (1) = 10$$

$$\Rightarrow f'(1) = \dfrac{10}{13}$$

#### If  $$(f(x))^2 + x\,f\,(x) = 6$$ and  $$f(1) = 2$$, find the value of $$f'(1).$$

A $$\dfrac{-2}{5}$$

B $$\dfrac{-3}{7}$$

C $$81$$

D $$\dfrac{4}{7}$$

×

Differentiate both  sides with respect to  $$x$$ treating $$f(x)$$  as a function of  $$x$$ .

$$\Rightarrow 2\,f(x) f' (x) + x\,f'(x) + f(x) = 0$$

Put  $$x=1$$  as  $$f'(1)$$  is the desired value .

$$\Rightarrow 2\,f(1) \,f' (1) + 1\,f'(1) +f(1) =0$$

$$\Rightarrow 2 × 2 × f'(1) + f' (1) + 2 = 0$$

$$\Rightarrow 5\,f' (1) +2=0$$

$$\Rightarrow f' (1) = \dfrac{-2}{5}$$

### If  $$(f(x))^2 + x\,f\,(x) = 6$$ and  $$f(1) = 2$$, find the value of $$f'(1).$$

A

$$\dfrac{-2}{5}$$

.

B

$$\dfrac{-3}{7}$$

C

$$81$$

D

$$\dfrac{4}{7}$$

Option A is Correct