# Trigonometric derivatives

• Ry122

#### Ry122

How do I get the derivative of
y=tan4x
The answer in the back of the textbook
says (4/cos^2(4X))

remember tan x can be written in terms of sin and cos

yes i know that tanx= sinx/cosx
y=tan4x -> y=sin4x/cos4x
Then what?

Also use the chain rule, u=4x.

y=sin4x/cos4x
u=4x
u'=4
y=sinu/cosu
y=cosu/-sinu
y=4cosu/-4sinu
y=4cos4x/-4sin4x
What am I doing wrong?

If you have a function f(x)=u/v then the derivative is f'(x)=$$\frac{u'v-v'u}{v^2}$$. So let u=sin x and v = cos x for this, you can find the derivative of tan x is (sec x)^2, then after that use the chain rule.

Using the quotient rule this is what i get.
u=sin4x
u'=4cos4x
v=cos4x
v'=-4sin4x

(-4sinX)(cos4X)--(4sinX)(sin4X)/(cos4X)^2

No no, forget about the 4x bit for now, just use the quotient rule with u=sin x and v=cos x to find the derivative of tan x.

One you know that, THEN use the chain rule. Do you know what the chain rule is?

Do you mind showing me the whole working out yourself?
That would help me alot.
Thanks

The video was more than sufficient.

If you still are unable to complete the problem, seek help from your teacher, peers, or get a tutor.

$$y=tan4x ; u=4x$$
$$\frac{dy}{dx}=\frac{d(tanu)}{du}.\frac{du}{dx}$$
$$\Rightarrow \ \frac{dy}{dx}= sec^2u. 4$$
$$\Rightarrow \ \frac{dy}{dx}=4sec^{2}4x$$

Last edited:
We haven't learned the other ratios yet.
The answer i need to get is in this form
(4/cos^2(4X))
I've finally figured out how to do this i just don't understand how
(4cos4x^2) + (4sin4x^2) / (4cos4x^2) = (4/cos^2(4X))
What happens to the cos and sin squared in the numerator?

Hint: isn't there a trig identity relating sin2 and cos2? too easy
thanks guys

We haven't learned the other ratios yet.
The answer i need to get is in this form
(4/cos^2(4X))
I've finally figured out how to do this i just don't understand how
(4cos4x^2) + (4sin4x^2) / (4cos4x^2) = (4/cos^2(4X))
What happens to the cos and sin squared in the numerator?
i thought you would know $\ sec^2x=\frac{1}{cos^2x}$ :uhh: