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

Question

Samantha Adkison · Accepted Answer

#-(sec(1/sqrt(x)) tan(1/sqrt(x)))/(2sqrt(x^3)) #

The chain rule is used whenever there are functions of functions. In other words, when you have a function of the form;

#f(x)=g(h(x))#

The chain rule tells us that the derivative of this function is;

#f'(x)=g'(h(x))*h'(x)#

The first step to using the chain rule is to identify the separate functions. Lets start by rewriting the given function to make the component functions easier to identify.

#1/cos(sqrt(1/x)) = cos^-1((1/x)^(1/2))=cos^-1(x^(-1/2))#

Now we have a function of the form;

#f(g(h(x)))#

Where;

#f(x)=x^-1# #g(x)=cosx# #h(x)=x^(-1/2)#

You can see that if you work backwards plugging these functions into each other, you get the original function back.

#f(g(h(x)))=f(g(x^(-1/2)))) = f(cos(x^(-1/2))) = cos^-1(x^(-1/2))#

We can take the derivative of each of these component functions to get;

#f'(x)=-x^-2# #g'(x)=-sinx# #h'(x)=-x^(-3/2)/2#

Now we apply the chain rule to our function.

#d/(dx)cos^-1(x^(-1/2))=-cos^-2(x^(-1/2))d/dx cos(x^(-1/2))#

Apply the chain rule again.

#=-cos^-2(x^(-1/2))(-sin(x^(-1/2)))d/dx x^(-1/2)#

Take the last derivative.

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

Simplify.

#=-(sec(1/sqrt(x)) tan(1/sqrt(x)))/(2sqrt(x^3)) #

Samantha Adkison · Answer

undefined To differentiate $ f(x) = \frac{1}{\cos(\sqrt{1/x})} $ using the chain rule, you need to recognize that it is composed of two functions: the outer function $ g(u) = \frac{1}{\cos(u)} $ and the inner function $ u(x) = \sqrt{1/x} $.

First, find the derivative of the inner function:

$$ u'(x) = \frac{d}{dx}(\sqrt{1/x}) $$

$$ = \frac{d}{dx}(x^{-1/2}) $$

$$ = -\frac{1}{2}x^{-3/2} $$

Next, find the derivative of the outer function with respect to its variable:

$$ g'(u) = \frac{d}{du}(\frac{1}{\cos(u)}) $$

$$ = -\frac{\sin(u)}{\cos^2(u)} $$

Now, apply the chain rule:

$$ \frac{d}{dx}(\frac{1}{\cos(\sqrt{1/x})}) = g'(u) \cdot u'(x) $$

$$ = -\frac{\sin(\sqrt{1/x})}{\cos^2(\sqrt{1/x})} \cdot (-\frac{1}{2}x^{-3/2}) $$

$$ = \frac{\sin(\sqrt{1/x})}{2x^{-3/2}\cos^2(\sqrt{1/x})} $$

So, the derivative of $ f(x) $ with respect to $ x $ using the chain rule is:

$$ \frac{\sin(\sqrt{1/x})}{2x^{-3/2}\cos^2(\sqrt{1/x})} $$