If #y=ln(sin(pi/2))#, then what is #dy/dx#?

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If #y=ln(sin(pi/2))#, then what is #dy/dx#?

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Answer 1 · 2016-12-19T14:05:47

0

Explanation:

Geometrical interpretation:

#y=ln sin (pi/2)=ln 1 = 0#. So, slope of this line, y'=(0)'=0

This equation represents the x-axis.

The slope of x-axis is 0.

Answer 2 · 2016-12-19T15:36:12

I tried using your corrected version Miroslav.

Explanation:

If you needed #y=ln(sin(pi/x))#
you need to use the Chain Rule: derive first the argument, then the sine and finally the log:
#(dy)/(dx)=(-pi/x^2)*cos(pi/x)*1/sin(pi/x)=-pi/x^2*cot(pi/x)#

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If #y=ln(sin(pi/2))#, then what is #dy/dx#?

2 Answers

Explanation:

Explanation:

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