Cheeky

Calculus Level 3

Let y = ln ⁡ ( 1 − x ) + z y = \ln(1-x) + z and x = sin ⁡ z x = \sin z . Find the value of d y d x \dfrac{dy}{dx} at x = 0 x=0 .

0 The function does not exist 2 1

Ritam Podder by Ritam Podder , 21, India

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2 solutions

J D
May 29, 2016

Substitute z=arcsinx, so we have y=ln(1-x)+arcsinx. Y'=-1/(1-x)+1/(root(1-x^2)). Y' at zero is -1+1=0

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Nanda Rahsyad
May 28, 2016

This is a very tricky question... it just mentioned d y d x \frac{dy}{dx} without mentioning what was being differentiated. So, there was literally N O F U N C T I O N \boxed{NO \boxed{FUNCTION}} whatsoever. Please correct me if I misunderstood the question, thank you. 😅

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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