# Thread: chain rule differentiation check!

1. ## chain rule differentiation check!

derivative of:

$\displaystyle y = sin(tan(sinx))$

chain rule:

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

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

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

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

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

then derivative of tan(sinx)

$\displaystyle sec^2(sinx) \cdot cosx$

$\displaystyle cos(tan(sinx)) \cdot sec^2(sinx) \cdot sec^2(sinx) \cdot cosx$
this correcT?

4. Nope but almost

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

5. Originally Posted by 11rdc11
Nope but almost

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