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Continue Shopping ```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 =esin2 x, find (d2 x)/dy2 in terms of x.

Solution:

Here y =esin2 x. Differentiating with respect to x, we get

dy/dx=sin 2x.esin2 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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