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y=tan(pi-(10/x)) 

My answer is (-10/x^2) *sec ^2(pi-(10/x))  but my other answer is  (10/x^2) *sec ^2(10/x) .  Not sure why the PI is not in the second answer ( the correct one) 

Guest May 3, 2018
 #2
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Find the derivative of the following via implicit differentiation:
d/dx(y) = d/dx(-tan(10/x))

The derivative of y is y'(x):
y'(x) = d/dx(-tan(10/x))

Factor out constants:
y'(x) = -d/dx(tan(10/x))

Using the chain rule, d/dx(tan(10/x)) = ( dtan(u))/( du) ( du)/( dx), where u = 10/x and d/( du)(tan(u)) = sec^2(u):
y'(x) = -d/dx(10/x) sec(10/x)^2

Factor out constants:
y'(x) = -sec^2(10/x) 10 d/dx(1/x)

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = -1.
d/dx(1/x) = d/dx(x^(-1)) = -x^(-2):
 
y'(x) = -10 sec^2(10/x) (-1)/(x^2) [Courtesy of Mathematica 11 Home Edition]

Guest May 3, 2018
 #3
avatar+93866 
+1

y=tan(pi-(10/x)) 

My answer is (-10/x^2) *sec ^2(pi-(10/x))  but my other answer is  (10/x^2) *sec ^2(10/x) .  Not sure why the PI is not in the second answer ( the correct one) 

 

\(y=tan(\pi-(10/x)) \\ y=tan(\frac{-10}{x}) \\\qquad\qquad x\ne0\\ \qquad\qquad \frac{-10}{x}\ne \frac{\pi}{2}\pm n\pi\\ \qquad\qquad\frac{-10}{x}\ne \frac{\pi\pm 2n\pi}{2}\\ \qquad\qquad\frac{x}{-10}\ne \frac{2}{\pi\pm 2n\pi}\\ \)

\(\qquad\qquad x\ne \frac{-20}{\pi\pm 2n\pi}\qquad x\ne 0 \qquad n\in Z\\\)

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\(y=tan(\frac{-10}{x}) \\ y=tan(-10x^{-1})\\ \frac{dy}{dx}=--10x^{-2}*sec^2(10x^{-1})\\ \frac{dy}{dx}=\frac{10}{x^2}*sec^2\frac{10}{x}\)

 

\(where \quad x\ne \frac{20}{\pi\pm 2n\pi}\qquad x\ne 0 \qquad n\in Z\)

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Melody  May 3, 2018
edited by Melody  May 3, 2018

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