is equal to
lt h->0 [sin(x+h)½ - sin(x)½]/h
Differentiate it with ‘h’
lt h->0 {cos(x+h)½ * ½(x+h)1/2] - 0}/1
lt h->0 {cos(x+h)½ * ½(x+h)1/2]
lt h->0 cos(x)½ / 2(x)½
Divide the numerator and denominator by x, so that given function becomes
f(x)=1+sinx/x/(1+cosx/x)
Now as x→∞.sinx/x→0 , because sin x would oscillate between +1 and -1, which in either case divided by ∞ would be 0. Thus the limit of the numerator would be 1. Like wise the limit of the denominator would also be 1.
Thus limit as a whole would be 1
is equal to
lim x→0 sin xn . (x)m . xn /(sin x)m.(x)m.xn
lim x→0 sin xn.(x)m.xn-m/xn.(sin x)m
Applying limits.
=0n-m = 0
If G(x) = then
has the value
then dy/dx is equal to
If f be a function such that f (9) = 9 and f ‘ (9) = 3, then is equal to
is equal to
If f(x) = , x ∈ (0,1), then f'(x) is equal to
If y = sin-11 x and z = cos -1 then dy/dz =
is equal to
The function, f(x) = and f(a) = 0, is
is equal to
is equal to
If sin x = then dx/dy is equal to
holds true for
Dervative of tan w.r.t
is
If y = log then
If y = log x , then yn =
The derivative of sec-1 with respect to
at x = 1/x is
is equal to
is eqaual to
If y = then dy/dx =
then at x = 1, f(x) is
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