if f(x)=|cosx-sinx| then find f'(π/6).
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Answer:
answer is 3/4. ... ... .. ..
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- For x =π/6, (cosx-sinx)=(√3/2-1/2)>0
- So the modulus opens without any sign change.
- f(x)=cosx-sinx
- f'(x)=-sinx-cosx
- f'(π/6)=-1/2-√3/2
=-1.366
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