Here's the question you clicked on:
onegirl
Find the derivative of f(x) tan 3x - csc 4x.
So you mean\[\frac{ d }{ dx }(\tan3x-4x)*\frac{ d }{ dx }(3x)*\frac{ d }{ dx }(4x)\]?
no the derivative of tan 3x - 4x...the multiplying part is the directions
hold on wait its tan 3x - csc 4x
its tan 3x- csc 4x i miss typed it
That's the answer you got or the correct answer?
And what part of the equation is inside the tan? Is it tan(3x)-4x or tan(3x-4x)
No not the correct answer i need to find the derivative of that and it says to multiply the derivatives of 3x and 4x into the promblem
its tan 3x - csc 4x thats how it written
@onegirl Please, post the right question. I confuse, too
let me edit my original question
there do you guys see it now?
yes. i think Twis7ed will guide you well.
Oh, ok, so\[\frac{ d }{ dx }(\tan 3x - \csc 4x)=\frac{ d }{ dx }\tan3x - \frac{ d }{ dx }\csc4x\] From that you know that \[\frac{ d }{ dx }\tan3x = 3\sec^2(3x)\] and that \[\frac{ d }{ dx }\csc4x=-4\cot(4x)\csc(4x)\]so you end up with\[\frac{ d }{ dx }(\tan3x-\csc4x)=3\sec^2(3x)+4\cot(4x)\csc(4x)\]
^Do you understand that?
yes, so i dont have to multiply the derivatives of 3x and 4x into the whole problem?
yes i understood what you wrote/
I'm not sure, what is the entire question?
Okay so I found the derivative of that and my teacher told me to multiply 3x and 4x into it here is how i did it hold on
d/dx (tan(3)x - csc 4x = tan(3) - csc 4
i think i understnad where i went wrong now after seeing what you showed
ALright, just try to remember that when you have a trig function you usually use the chain rule with the function and then the function inside of it and it should be easy :)