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