# Thread: Find the derivative of the function y = tan...

1. ## Find the derivative of the function y = tan...

Find the derivative of the function.

y = tan(9θ)^2

y'(θ) = ?

2. Originally Posted by question111
Find the derivative of the function.

y = tan(9θ)^2

y'(θ) = ?
by the chain rule: $\displaystyle \frac d{dx} \tan u = u' \sec^2 u$ where $\displaystyle u$ is a function of $\displaystyle x$

does that help?

3. Mmm, not exactly. I'm quite behind in my work, and I don't really see how to use the info you gave me there...

I'm particularly confused by the symbol θ ?
I honestly don't know what that symbol is supposed to mean for this equation.

4. the symbol looks unclear in the problem but is it suppose to look like a zero with a horizontal line in it? if so its just a variable. it is like having x in there.

5. Originally Posted by question111
Find the derivative of the function.

y = tan(9θ)^2

y'(θ) = ?
$\displaystyle y = \tan^2{\theta} = (\tan{\theta})^2$.

Let $\displaystyle u = \tan{\theta}$, then $\displaystyle y = u^2$.

Now use the chain rule, $\displaystyle \frac{dy}{d\theta} = \frac{du}{d\theta}\times \frac{dy}{du}$.

$\displaystyle \frac{du}{d\theta} = \sec^2{\theta}$

$\displaystyle \frac{dy}{du} = 2u = 2\tan{\theta}$

So $\displaystyle \frac{dy}{d\theta} = 2\tan{\theta}\sec^2{\theta}$.

Hope that helped.

6. Originally Posted by Prove It
$\displaystyle y = \tan^2{\theta} = (\tan{\theta})^2$.

Let $\displaystyle u = \tan{\theta}$, then $\displaystyle y = u^2$.

Now use the chain rule, $\displaystyle \frac{dy}{d\theta} = \frac{du}{d\theta}\times \frac{dy}{du}$.

$\displaystyle \frac{du}{d\theta} = \sec^2{\theta}$

$\displaystyle \frac{dy}{du} = 2u = 2\tan{\theta}$

So $\displaystyle \frac{dy}{d\theta} = 2\tan{\theta}\sec^2{\theta}$.

Hope that helped.
the OP seemed to write $\displaystyle y = \tan (9 \theta)^2$ as opposed to $\displaystyle \tan^2 9 \theta$. i wonder if it was a typo....?