# Math Help - chain rule differentiation check!

1. ## chain rule differentiation check!

derivative of:

$

y = sin(tan(sinx))
$

chain rule:

$

y' = cos(cosxsec^2x(sinx))
$

i got upto there but im ment to differntiate this but im not sure what to do inside that bracket to differentiate that.

2. $y'=\cos (\tan (\sin x))\cdot \left( \tan (\sin x) \right)',$ and chain rule again.

3. Originally Posted by Krizalid
$y'=\cos (\tan (\sin x))\cdot \left( \tan (\sin x) \right)',$ and chain rule again.
hrmm not to sure what u mean. first i take the derivative of sin(tan(sinx))
which is:

$

cos(tan(sinx)) \cdot sec^2(sinx)

$

then derivative of tan(sinx)

$

sec^2(sinx) \cdot cosx

$

$

cos(tan(sinx)) \cdot sec^2(sinx) \cdot sec^2(sinx) \cdot cosx

$

this correcT?

4. Nope but almost

$cos(tan(sinx)) \cdot sec^2(sinx) \cdot cosx$

5. Originally Posted by 11rdc11
Nope but almost

$cos(tan(sinx)) \cdot sec^2(sinx) \cdot cosx$
thank you