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.

Related questions
See all questions in Product Rule
Impact of this question
3818 views around the world
You can reuse this answer
Creative Commons License