How do you differentiate #y=x^2tan(1/x)#?

1 Answer
Geoff K.
Nov 19, 2016

Use a combination of the product rule and the chain rule to get
#dy/dx=2xtan(1/x)-sec^2(1/x)#.

Explanation:

Let #f# and #g# be functions of #x#.
The product rule states that if #y = f*g# then #y'=f'*g+f*g'#.
The chain rule states that if #y=f(g)# then #y'=f'(g)*g'#.

In this question, say #f(x)=x^2#, #g(x)=tan(x)#, and #h(x)=1/x#.

Then #y=f(x)*g[h(x)]#

And so
#y'=[f(x)]'g[h(x)]+f(x){g[h(x)]}'#
#y'=f'(x)*g[h(x)]+f(x)*g'[h(x)]*h'(x)#

This means if #y=x^2tan(1/x)#
then

#dy/dx=d/dx(x^2)*tan(1/x)+x^2*d/dxtan(1/x)#

#dy/dx=2xtan(1/x)+x^2*sec^2(1/x)*d/dx(1/x)#

#dy/dx=2xtan(1/x)+cancel(x^2)sec^2(1/x)*(-1)/cancel(x^2)#

#dy/dx=2xtan(1/x)-sec^2(1/x)#

That's likely as far as you'll need to go.

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