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

2 Answers
Harish Chandra Rajpoot
Jul 18, 2018

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

Explanation:

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}

maganbhai P.
Jul 18, 2018

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

Explanation:

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

Using Chain Rule :

(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))

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