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

Illustration Questions

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.

Illustration Questions

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 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  \(y-y_1=y' (x-x_1)\)

Illustration Questions

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 \( x\).

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

Illustration Questions

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

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.

Illustration Questions

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

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 \(m_1\,m_2=-1\))

   

Illustration Questions

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  \(x=1\) .

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

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

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

Illustration Questions

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

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 \(y\) .

Illustration Questions

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

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