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Continue Shopping ```General Theorems on Differentiation

General Theorems on Differentiation d/dx (c) = 0

d/dx [a f(x)+b g(x) ] = af'(x) + b g'(x)

d/dx [f(x)g(x) ] = f' (x)g(x) + f(x) g'(x)

d/dx [f(x)/g(x)] = (g(x) f'(x) - f(x) g'(x))/[g(x) ]2

d/dx [f(x)g(x) ] = f(x)g(x) [g(x)/f(x) f'(x) + g' (x)lnf(x)]

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)

Illustration:

Differentiate

y = sec-1 by ab- nitio

sec y=x                              ...... (i)

Let Δx be increment in x and Δy be the corresponding increment in y

x + Δx = sec (y+Δy)                ...... (ii)

(Equation (ii)-Equation (i)) gives

Δx = sec (y + Δy)- sec y

Δx/Δy=(sec (y+?y)- secy)/(? y)

Applying limits Δ y-->0

lim?y→0 ?x/?y=lim?y→0 ( sec (y+?y)-secy)/(? y) (0/0 form)

dx/dy=lim?y→0 (2 sin?y/2 sin(y+?y/2) )/(?y.cosy cos(y+?y) )

=lim?y→0 (sin?y/s/(?y/2)) × lim?y→0 sin(y+?y/2)/cosy cos(y+?y)

--> (dx )/(dy ) = siny/(cos2 y)

--> (dx )/(dy ) = 1/(dx/dy) =(cos2 y)/siny

= 1/tany secy =1/(x √(x2-1)) (wrong)

sec y = x                   (Given)

1+ tan2y = sec2 y

tan y = ± √(sec2 y-1)

= ± √(x2 -1)

Sec-1 x = y ? (0, Π)

--> (dx )/(dy ) = 1/(|x| √(x2-1))

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