# Math Help - sin cos tan derivatives

1. ## sin cos tan derivatives

Hey guys im stuck on 3 problems that are due in 2 days. I am hoping these problems are super easy for you guys to do so I can learn from the steps posted and ace my test 3 days from now. So here are the 3 problems.

Use the chain rule (fog) = f(g(x)) to find the derivative dy/dx of each of the functions below.

1. y= x*tan(square)x / 1+secx

2. y=sin(sinx)

3. y=cos(tanx)

These are the problems that can help me past the test so please help me out.
THANKS A LOT!!!

2. tell us what is it that you don't understand.

if you know how to apply the chain rule, and you are familiar with the the basic derivatives of the trigonometric functions (cos, sin tan...) there is no problem finding those derivatives.

$

\frac{d}
{{dx}}f\left( {g(x)} \right) = \left. {\frac{d}
{{dt}}f(t)} \right|_{t = g(x)} \frac{d}
{{dx}}g(x)

$

3. i know the rules but i don't know how to apply them to the problem to come up with the answer.

4. Originally Posted by uniquereason81
i know the rules but i don't know how to apply them to the problem to come up with the answer.
for exmple:

2. f(x) = sinx, g(x) = sinx

3. f(x) = cosx g(x) = tanx

now can you solve 2 & 3 ?

5. thanks Peritus but can u go futher with the problems. I understand how you got f(x) and g(x). THANKS

6. Originally Posted by uniquereason81
thanks Peritus but can u go futher with the problems. I understand how you got f(x) and g(x). THANKS
can you apply the chain rule:

$
\frac{d}
{{dx}}f\left( {g(x)} \right) = \left. {\frac{d}
{{dt}}f(t)} \right|_{t = g(x)} \frac{d}
{{dx}}g(x)
$

to f(x) and g(x) in 2& 3 ?

7. no i can't apply it. I'm sorry I just learned this stuff last week.

8. Originally Posted by uniquereason81
no i can't apply it. I'm sorry I just learned this stuff last week.
ok, let us work problem 3 step by step:

$\begin{gathered}
f(x) = \cos (x) \hfill \\
g(x) = \tan (x) \hfill \\
\end{gathered}$

now let's look at the chain rule:

$
\frac{d}
{{dx}}f\left( {g(x)} \right) = \left. {\frac{d}
{{dt}}f(t)} \right|_{t = g(x)} \frac{d}
{{dx}}g(x)
$

first we have to differentiate f(x):

$f'(x) = - \sin (x)$

now we have to substitute g(x) as the argument of this result:

$
- \sin (\tan x)
$

the next step is to calculate the derivative of g(x):

$
g'(x) = \frac{1}
{{\cos ^2 (x)}}$

the derivative of this composite function is the product of these two results:

$\frac{d}
{{dx}}\cos (\tan x) = \frac{{ - \sin (\tan x)}}
{{\cos ^2 (x)}}$

9. okay i think i got it so thanks but one question how do you know the inverses of sin cos and tan