Question

For x∈R, f(x) = |log2 – sinx| and g(x) = f(f(x)), then:
(1) g is differentiable at x = 0 and g'(0) = –sin(log2)
(2) g is not differentiable at x = 0
(3) g'(0) = cos(log2)
(4) g'(0) = –cos(log2)

Answer 1

Answer:

f(x)=|log2−sinx|

g(x)=f(f(x))=|log2−sin(f(x))|

g(x)=|log2−sin(|log2−sinx|)|

g'(x)=|0−cos(|log2−sinx|)×(0−cosx)|

=|+cos(log2−sinx∣)×cosx|

g'(0)=|cos(log2)|

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Step-by-step explanation:

Answer 2

Step-by-step explanation:

Hello Dear

we have f(x) = |log2 - sinx|. { given

for x→0 log2 > sin x

since log 2 = 0.301

Therefore f(x) = log2 - sinx. { |x| = x if x>0

g(x) = log2 - sin(f(x))

=> log2 - sin(log2 -sinx)

Now from here we can conclude that g(x) is differentiable at x = 0 since sinx is diff. at x

so g'(x) =. 0 -cos(log2 -sinx)(-cosx)

=. cosx(cos(log2 - sinx))

so

g'(0) = cos(log2)

hence

option 2 is correct.