How do you differentiate #y=2^(arcsin(sqrtt))#?

1 Answer
Apr 18, 2018

#dy/dt=(2^arcsinsqrttln2)/(2sqrt(t-t^2)#

Explanation:

We can use logarithmic differentiation:

#y=2^arcsin(sqrtt)#

Apply the logarithm to both sides:

#lny=ln(2^arcsinsqrtt)#

Recalling that #ln(a^b)=blna,# we get

#lny=arcsinsqrttln2#

Now, differentiate both sides with respect to #t#. This will result in implicitly differentiating the left side. Furthermore, recall that #d/dxarcsinx=1/sqrt(1-x^2)#

#1/y*dy/dt=ln2/sqrt(1-(sqrtt)^2)*d/dtsqrtt#

#1/y*dy/dt=ln2/(2sqrt(t-t^2)#

Solve for #dy/dt:#

#dy/dt=y*ln2/(2sqrt(t-t^2)#

#dy/dt=(2^arcsinsqrttln2)/(2sqrt(t-t^2)#