MHB Derivatives of trigonometric functions

[jhe2]

The answer for derivative of y=5tanx+4cotx is y'=-5cscx^2. But how come on math help the answer is 5sec^2x-4csc^2x? I have a calculus test coming up and I really would appreciate if someone could explain!

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Oh nvm I see my mistake!

 

[MarkFL]

Hello, and welcome to MHB! (Wave)

The second answer you cited agrees with what W|A outputs as the derivative for the given function. If the two answers are equivalent, then the following will be an identity (I am assuming you mean something a bit different from the first answer):

$$-5\csc^2(x)=5\sec^2(x)-4\csc^2(x)$$

$$-\csc^2(x)=5\sec^2(x)$$

This is not an identity, so the first result you posted isn't correct. Where did this first result come from?