# Differentiating tan2x

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#1
Hello,

On paper it says that the answer to differentiating tan2x is 2(sec(x))^2

Can someone explain to me how to get this answer? My answer is (sec(2x))^2 and I've got no clue where they get the 2 from..
1
8 years ago
#2
Have you done the chain rule?
0
8 years ago
#3
Standard chain rule. Make a substitution u = 2x and it should be obvious.
0
8 years ago
#4
Differentiating tan2x should give 2sec(2x)^2.
Such type of differentiation is explained in C3.

The general way of differentiating such an expression, tan(y), is to differentiate the part in bracket and multiply it with sec(y)^2.

Differentiating 2x gives you 2. So your differentiated expression should be 2*sec(2x)^2.
Last edited by RK; 4 months ago
1
3 years ago
#5
yea rememba tanx is sinx/cosx by identity, my own ans is 2[sec^2X] 2
0
3 years ago
#6
(Original post by lokoabe2018)
yea rememba tanx is sinx/cosx by identity, my own ans is 2[sec^2X] 2
(Original post by an_atheist)
if we sub 2x=u, we get y=tanu,u=2x
dy/du=sec^2(u),du/dx=2
dy/dx=dy/du * du/dx (chain rule)
=sec^2(u) * 2
=2sec^2(2x) (sub u=2x back in)
5 years late.
1
3 years ago
#7
(Original post by RDKGames)
5 years late.
So we are. That'll learn me to check the timestamp on the OP.
0
3 years ago
#8
You must use the chain rule to differentiate this. i.e. multiplying the derivatives of of each function. There are "two functions" within tan2x. The tan, and the 2x.
To differentiate the tan, you get sec^2. That's that sorted. You then must multiply this by the derivative of the 2x, which is 2. You keep the 2x after the trig function, no matter how many times it is differentiated. So, you have 2sec^2(2x).
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