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L'Hopital's rules says that the
lim x→a f(x)/g(x)
⇒ f'(a)/g'(a)
Using this, we get
lim x→0 (1−cosx)/x^{2}
⇒ − sin0/2(0)
Yet as the denominator is 0, this is impossible. So we do a second limit:
lim(x→0) sinx/2x
⇒ cos0/2 = 1/2 = 0.5
So, in total lim x→0 (1−cosx)/x^{2}
⇒ lim x→0 sinx/2x
⇒ cosx/2
⇒ cos0/2= 1/2
If f (x) is a polynomial of degree m (⩾1) , then which of the following is not true ?
As all the three remaining statements are true for the given function.
Let f and g be differentiable functions such that fog = I, the identity function. If g’ (a) = 2 and g (a) = b, then f ‘ (b) =
f(g(x)) = x
f'(g(x)) g'(x) = 1
put x = a
f'(b) g'(a) = 1
2 f'(b) = 1
f'(b) = 1/2
If f (x) =x^{2}g(x) and g (x) is twice differentiable then f’’’ (x) is equal to
The graph of f(x) = x
As observed from the graph, f(x) = x is continuous at x = 0.
As this curve is pointed at x = 0, f(x) is not derivable at x = 0.
Differential coefficient of a function f (g (x)) w.r.t. the function g (x) is
Given x^{p}y^{q} = (x+y)^{p+q}Taking log on both sides we get:
_{p}logx+_{q}logy = (p+q) log (x+y). Differentiating both sides w.r.t. x we get ,
If y = ae^{mx} + be^{−mx}, then y_{2} is equal to
y = ae^{mx} + be^{mx} ⇒ y1 = ame^{mx} + (m)be^{mx} ⇒y2
= am^{2}e^{mx}_{ + (m}^{2})be^{mx} ⇒y2 = m^{2} (ae^{mx} + be^{mx}) ⇒ y2 = m^{2}y
If f(x) be any function which assumes only positive values and f’ (x) exists then f’ (x) is equal to
If y = a sin mx + b cos m x, then is equal to
y = a sin mx+b cos mx ⇒ y_{1} = am cos mx − bm sin mx
⇒ y_{2} = −am^{2}sin mx−bm^{2}cosmx ⇒ y_{2} = −m^{2}(a sin mx+b cos mx)
If y = tan^{−1}x and z = cot^{−1}x then is equal to
If both f and g are defined in a nhd of 0 ; f(0) = 0 = g(0) and f ‘ (0) = 8 = g’ (0), then equal to
(by using L’Hospital Rule)
The differential coefficient of log ( log x ) w.r.t. log x is
If f is derivable at x = a , then is equal to
209 videos218 docs139 tests

209 videos218 docs139 tests
