How do you differentiate #f(x)=1/sin(sqrt(1/x))# using the chain rule?

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Answer 1 · 2015-12-30T05:49:51

#f'(x)=(cos(1/sqrtx))/(2sin^2(1/sqrtx)x^(3/2))#

#f(x)=(sin(x^(-1/2)))^-1#

Use the chain rule: #d/dxu^-1=-u^-2*(du)/dx#

#f'(x)=-sin^-2(x^(-1/2))*color(blue)(d/dxsin(x^(-1/2))#

To find this derivative, use chain rule again:

#d/dxsin(u)=cos(u)*(du)/dx#

Thus,

#color(blue)(d/dxsin(x^(-1/2)))=cos(x^(-1/2))*d/dx x^(-1/2)#

#=>color(blue)(-1/2x^(-3/2)cos(x^(-1/2))#

Plug this back in.

#f'(x)=-sin^-2(x^(-1/2))*-1/2x^(-3/2)cos(x^(-1/2))#

#=>(cos(1/sqrtx))/(2sin^2(1/sqrtx)x^(3/2))#