How do you find the derivative of y=tan(2x)y=tan(2x)?
3 Answers
Essentially we use the rule:
d/dx(a* tan(k*x+h))=(a*k)/(cos^2(k*x+h))ddx(a⋅tan(k⋅x+h))=a⋅kcos2(k⋅x+h)
or
a*k*sec^2(k*x+h)a⋅k⋅sec2(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)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
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