How do you find the derivative of #y=tan(2x)#?
3 Answers
Essentially we use the rule:
#d/dx(a* tan(k*x+h))=(a*k)/(cos^2(k*x+h))#
or
#a*k*sec^2(k*x+h)#
I'll go through the steps behind it as well:
Derivitave of
The first step is to separate it into
#y=sin(2x)/cos(2x)#
Then we integrate using the quotient rule:
#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
Thus:
#y'=(2(cos^2(2x)+sin^2(2x)))/(cos^2(2x))#
becomes
#y'=(2)/(cos^2(2x))#
#=2sec^2(2x)#
There is a simple answer to this.
Thus, with
We can Use Chain rule for this.
Explanation:
We have to find derivative of
Let us assume
Then
By Chain Rule of Differenciation
Substitute u=2x in Eqn 1
Therefore