Theorems on Differentiation - Study Material for IIT JEE

Calculus
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General Theorems on Differentiation

 

General Theorems on Differentiation

 



Chain Rule

If y = f(u) and u = g(x), then dy/dx = dy/dx.du/dx = f' (g(x) ) g' (x)

e.g. Let y = [f(x)]ⁿ. We put u = f(x). so that y = uⁿ.

                Therefore, using chain rule, we get

                dy/dx = dy/dx.du/dx = nu^(n-1)   [f' (x)]^(n-1) f' (x)



Differentiation of parametrically defined functions 

 If x and y are functions of parameter t, first find dx/dt and dy/dt separately.

 Then dy/dx=(dy/dt)/(dx/dt)

e.g., x=a(Θ + sin Θ), y = a(1-cos Θ) where Θ is a parameter.

dy/dx = (dy/dΘ)/(dx/dΘ) = (a sinΘ)/(a (1+cos Θ))

         = (2sin Θ/2 cos Θ/2 )/(2 cos² Θ/2 ) = tan Θ/2

Higher Order Derivatives

(d² y)/dx² - d/dx (dy/dx),   (d³ y)/dx³  = d/dx ((d² y)/(dx² ))

(dⁿ y)/dxⁿ -d/dx ((d^(n-1) y)/dx^(n-1) );   (dⁿ y)/dxⁿ  

is called the nth order derivative of y with respect to x.

Illustration:

        If y = (sin^-1x)²+ k sin^-1x, show that (1-x²) (d² y)/dx² - x dy/dx = 2

Solution:  

          Here y = (sin^-1x)²+ k sin^-1x.

       Differentiating both sides with respect to x, we have

     Dy/dx = 2(sin^-1 x)/√(1-x² ) + k/√(1-x² )

⇒(1-x² ) (dy/dx)² = 4y + k²

      Differentiating this with respect to x, we get

       (1-x²) 2 dy/dx.(d² y)/(dx² ) - 2x (dy/dx)² = 4(dy/dx)

⇒(1-x² ) ( d² y)/dx² -x dy/dx = 2

  Illustration:     

If y =e^sin2^x, find (d² x)/dy² in terms of x.

Solution:            

Here y =e^sin2^x. Differentiating with respect to x, we get

 dy/dx=sin 2x.e^{sin2 x}  ⇒dx/dy = cosec 2x.e^{-sin2 x}

 Differentiating with respect to y, we get

  (d² x)/dy² = d/dy (cosec 2x.e^{-sin2 x} ) = d/dx (cosec 2x.e^{- sin2 x} ) dx/dy

             = (-2 cosec 2xcot 2x e^{-sin2 x}- e^{-sin2 x})

            = (-2 cosec² 2x cot 2x + cosec 2x) e^{-sin2 x}

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