How do you differentiate #f(x) = sin(sqrt(arcsinx)) # using the chain rule?

1 Answer
Jan 25, 2018

Answer:

#f'(x)=d/dx[sin(sqrtarcsinx)]=cos(sqrtarcsinx)/(2sqrtarcsinxsqrt(1-x^2))#

Explanation:

Given: #f(x)=sin(sqrtarcsinx)#

We'll use the chain rule but to make this easier to visualize we are going to let #sqrt(arcsinx)=u#

So we have

#f(x)=sin(u)#

By the chain rule, we are solving

#f'(x)=d/dx[sin(sqrtarcsinx)]=color(red)((df)/(du)[sin(u)])*color(blue)((du)/dx[sqrtarcsinx]#

The derivative of #color(red)(sin(u)# is #color(red)(cos(u))#

#:.color(red)((df)/dx[sin(u)]=cos(u)#

For #color(blue)((du)/dx[sqrtarcsinx]# we can use the chain rule again.

We can call #sqrtarcsinx# a new function, let's call it #g(x)#

#g(x)=sqrtarcsinx=>color(blue)(g'(x)=(dg)/dx[sqrtarcsinx]=(du)/dx[sqrtarcsinx]#

And similarly we'll let #arcsinx=v#

So we have

#g(x)=sqrtv#

By the chain rule we are differentiating

#color(blue)((dg)/dx[sqrtarcsinx])=color(purple)((dg)/(dv)[sqrtv]*color(green)((dv)/dx[arcsinx]#

For #color(purple)((dg)/(dv)[sqrtv]=>d/dx[v^(1/2)]#

By the Power Rule

#color(purple)((dg)/(dv)=1/2v^(1/2-1)=1/2v^(-1/2)=1/(2v^(1/2))=1/(2sqrtv)#

#:.color(purple)((dg)/(dv)[sqrtv]=1/(2sqrtv)#

For #color(green)((dv)/dx[arcsinx]#, it's derivative is #color(green)(1/sqrt(1-x^2)#

#:.color(green)((dv)/dx[arcsinx]=1/sqrt(1-x^2)#

Now that we know what #color(purple)((dg)/(dv) and color(green)((dv)/dx# are, we can find #color(blue)((dg)/dx[sqrtarcsinx])#

#color(blue)((dg)/dx[sqrtarcsinx])=color(purple)((dg)/(dv)[sqrtv])*color(green)((dv)/dx[arcsinx])=1/(2sqrtv)*1/sqrt(1-x^2)=1/(2sqrtvsqrt(1-x^2))#

Okay, now let us summarize

#u=sqrtarcsinx " " v=arcsinx#

#color(red)((df)/dx[sin(u)]=cos(u) " " color(blue)((du)/dx[sqrtarcsinx]=1/(2sqrtvsqrt(1-x^2)))#

Going back to the original problem:

#f'(x)=color(red)((df)/(du)[sin(u)])*color(blue)((du)/dx[sqrtarcsinx])=color(red)(cos(u))*color(blue)(1/(2sqrtvsqrt(1-x^2)))#

Now we have to substitute in back the values for #u# and #v# which will give us the final result:

#f'(x)=color(red)(cos(sqrtarcsinx))*color(blue)(1/(2sqrtarcsinxsqrt(1-x^2)))#

#f'(x)=color(red)(cos(sqrtarcsinx))/color(blue)((2sqrtarcsinxsqrt(1-x^2)))#

I hope this helped! :)