How do you find the derivative of y=tan(2x)y=tan(2x)?

3 Answers
Aug 5, 2014

Essentially we use the rule:

d/dx(a* tan(k*x+h))=(a*k)/(cos^2(k*x+h))ddx(atan(kx+h))=akcos2(kx+h)

or

a*k*sec^2(k*x+h)aksec2(kx+h)

I'll go through the steps behind it as well:

Derivitave of y=tan(2x)y=tan(2x)

The first step is to separate it into sin(x)sin(x) and cos(x)cos(x):

y=sin(2x)/cos(2x)y=sin(2x)cos(2x)

Then we integrate using the quotient rule:

f(x)=g(x)/(h(x)),f(x)=g(x)h(x), f'(x)= (g'(x)*h(x)-h'(x)*g(x))/(h(x))^2

y'=(cos(2x)xx2cos(2x) - sin(2x)xx-2sin(2x))/(cos^2(2x))

y'=(2cos^2(2x)+2sin^2(2x))/(cos^2(2x))

y'=(2(cos^2(2x)+sin^2(2x)))/(cos^2(2x))

To simplify this further we use the Pythagorean Identity:

sin^2(x)+cos^2(x)=1

So 2(cos^2(2x)+sin^2(2x)) becomes 2xx1=2

Thus:

y'=(2(cos^2(2x)+sin^2(2x)))/(cos^2(2x))

becomes

y'=(2)/(cos^2(2x))

=2sec^2(2x)

Jun 20, 2015

There is a simple answer to this.

d/(dx)[tanu] = sec^2u((du)/(dx))

Thus, with u = 2x (thus, u/x = 2), (du)/(dx) = 2, and

d/(dx)[tan(2x)] = sec^2(2x)*2 = color(blue)(2sec^2(2x))

We can Use Chain rule for this.

Explanation:

We have to find derivative of color(blue)(y=tan(2x))
Let us assume u=2x
Then
color(blue)(y=tanu & color(blue)(u=2x
By Chain Rule of Differenciation
dy/dx=dy/(du)*(du)/(dx)
d/(dx)(tan2x)=(d/(du)(tanu))(d/dx(2x))
d/(dx)(tan2x)=sec^2u*2 ->1
Substitute u=2x in Eqn 1
Therefore
d/dx(tan2x)=2sec^2 2x