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

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Answer 1 · 2018-07-18T11:55:13

#f'(x)={\sec(\sqrt\ln x)\tan(\sqrt\ln x)}/{2x\sqrt\ln x}#

Given that

#f(x)=1/{\cos\sqrt{\ln x}}#

#f(x)=\sec\sqrt{\ln x}#

Differentiating w.r.t. #x# using chain rule as follows

#f(x)=\sec\sqrt{\ln x}#

#=d/dx(\sec(\sqrt{\ln x}))#

#=\sec(\sqrt\ln x)\tan(\sqrt\ln x)\frac{d}{dx}(\sqrt\ln x)#

#=\sec(\sqrt\ln x)\tan(\sqrt\ln x)\frac{1}{2\sqrtln x}d/dx\ln x#

#=\sec(\sqrt\ln x)\tan(\sqrt\ln x)\frac{1}{2\sqrtln x}1/x#

#={\sec(\sqrt\ln x)\tan(\sqrt\ln x)}/{2x\sqrt\ln x}#

Answer 2 · 2018-07-18T11:58:26

#f'(x)=1/(2xsqrt(lnx))sec(sqrt(lnx))tan(sqrt(lnx))#

Here ,

#f(x)=y=1/cossqrtlnx#

Let ,

#y=1/u# , #u=cosv# , #v=sqrtw# , #w=lnx#

So,

#(dy)/(du)=-1/u^2# , #(du)/(dv)=-sinv# , # (dv)/(dw)=1/(2sqrtw)and(dw)/(dx)=1/x#

#(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dw)(dw)/(dx)#

#:.(dy)/(dx)=(-1/u^2)(-sinv)(1/(2sqrtw))(1/x)#

#:.(dy)/(dx)=1/cos^2v*sinv(1/(2sqrtw))(1/x)to[becauseu=cosv]#

#=>(dy)/(dx)=sin(sqrtw)/cos^2(sqrtw)*1/(2sqrtw) (1/x)to[becausev=sqrtw]#

#=>(dy)/(dx)=sin(sqrt(lnx))/cos^2(sqrt(lnx))1/(2sqrt(lnx)) (1/x)to[becausew=lnx]#

Hence,

#f'(x)=1/(2xsqrt(lnx))sin(sqrtlnx)/cos^2(sqrt(lnx)) or#

#f'(x)=1/(2xsqrt(lnx))sec(sqrt(lnx))tan(sqrt(lnx))#