Definition Of A Derivative And Degrees

Calculus Level 2

Above shows the derivative of sin ⁡ ( x ) \sin(x) by the first principle.

What's the derivative of sin ⁡ ( x ∘ ) \sin(x^{\circ}) ?

π 180 cos ⁡ ( x ∘ ) \frac{\pi}{180}\cos(x^{\circ}) π 180 cos ⁡ ( x ) \frac{\pi}{180}\cos(x) cos ⁡ ( x ) \cos(x) cos ⁡ ( x ∘ ) \cos(x^{\circ})

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Lokesh Sharma by Lokesh Sharma , 25, India

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3 solutions

Garrett Clarke
Jun 9, 2015

The key to this problem is knowing that radians = (degrees * π)÷180.

If you have to take the derivative of sin(x) where x is in degrees, you're really taking the derivative of sin((π÷180) * x). Using the chain rule we have (π÷180) * cos((π÷180) * x), where (π÷180) * x is really x taken as degrees instead of radians, giving you the answer shown above.

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I did not regard x^0 as x in degrees, but thought it was a trick quesyin in that x^0 = 1. Edwin gray

Edwin Gray - 4 years, 1 month ago
Sarthak Tanwani
Jun 10, 2015

The derivation given above can also be used. In the last step limit as h approaches to zero sin(h)/h is actually π/180 if h is in degrees.

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Rahul Saxena
Oct 7, 2015

what was that head banging all about :)

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