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How do you find the derivative of #y=1+tan^2(5x)#?

Answer 1

Elijah Abram

#y'=10sin(5x)/cos^3(5x)#

Since the derivative of a constant function is null,

and the derivative of #tan(color(blue)f(x))# is

#1/cos^2color(blue)(f(x))*f'(x)#

and the derivative of #(color(red)(g(x)^2))# is #color(red)(2g(x))*g'(x)#,

then

#y'=0+color(red)(2tan(5x))*1/cos^2(color(blue)(5x))*5#

#=10sin(5x)/cos(5x)*1/cos^2(5x)#

#=10sin(5x)/cos^3(5x)#

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Answer 2

Axel Alvarez

To find the derivative of ( y = 1 + \tan^2(5x) ), we can use the chain rule and the derivative of (\tan^2(u)), where ( u = 5x ).

[ \begin{aligned} \frac{dy}{dx} &= \frac{d}{dx} \left(1 + \tan^2(5x)\right) \ &= \frac{d}{dx} \left(1 + \tan^2(u)\right) \quad \text{where } u = 5x \ &= \frac{d}{du} \left(1 + \tan^2(u)\right) \cdot \frac{du}{dx} \ &= 2\tan(u) \cdot \sec^2(u) \cdot 5 \ &= 10\tan(5x) \cdot \sec^2(5x) \end{aligned} ]

Therefore, the derivative of ( y = 1 + \tan^2(5x) ) with respect to ( x ) is ( 10\tan(5x) \cdot \sec^2(5x) ).

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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.