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How do you differentiate #y = (x + tan 2pi x )^2#?

Answer 1

William Abdalla

#y'=2(x+tan(2pix))(1+2pi*(1+tan^2(2pix))#

We employ that

#(tan(x))'=1/cos^2(x)=(sin^2(x)+cos^2(x))/cos^2(x)=1+tan^2(x)# and (x^n)'=nx^(n-1)# and the chain rule:

#(f(g(x)))'=f'(g(x))*g'(x)# so we get

#y'=2(x+tan(2pix))(1+2pi(1+tan^2(2pix)))#

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Answer 2

Cameron Alexander

To differentiate ( y = (x + \tan(2\pi x))^2 ), you can use the chain rule along with the power rule. First, differentiate the inner function ( x + \tan(2\pi x) ) with respect to ( x ), then square the result.

Using the chain rule, the derivative of ( x + \tan(2\pi x) ) with respect to ( x ) is:

[ \frac{d}{dx}(x + \tan(2\pi x)) = 1 + \sec^2(2\pi x) \cdot 2\pi ]

Then, square the result:

[ \left(1 + \sec^2(2\pi x) \cdot 2\pi \right)^2 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.