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

Since the derivative of a constant function is null,

then

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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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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.

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.

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.

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