IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

v_gurucharan

diff.

neeraj_agarwal_1990

1) f'(x) = |cosx|/cosx . d(cosx)/dx (using the result and chain rule)

so f'(x) = -|cosx|tanx
f'(3pi/4) = +1/root2

2) similar

3)f'(x)=|ln|x||/ln|x| . d(ln|x|)/dx (using chain rule)

=> f'(x)= |ln|x||/ln|x| .1/|x| .d|x|/dx

=> f'(x)= |ln|x||/ln|x| .1/|x| . |x|/x

put the desired value of x..

spideyunlimited

1 b, 2c, 3a

spideyunlimited

for 1)
the given value of x is positive
so f(x) will be + cos x
and f'(x) will be -sinx.
put x = 3pi/4
you get 1/root2

so 1 --> (b)

2) here value of x is negative
so f(x) will be sin(-x) = -sinx
and f'(x) = -cosx
put x = -pi/6
you get - root3/2

so 2 --> (c)

3) here value of x is -1/2
= -2^(-1)

so f(x) = - ln (-x)
f'(x) = 1/x
put x = -1/2
we get -2

so 3 --> (a)

varshavallig

1)1/root2
2)-root 3/2
3)2
!!!!!!!!!!!!!

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