How do you use the chain rule to differentiate #y=x^2tan(1/x)#?

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Answer 1 · 2017-01-26T11:08:52

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

You need the chain rule and the product rule here:

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

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

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

Now we use the chain rule with #y=1/x# to calculate:

#d/(dx) tan(1/x) = (d tany)/dy * (dy)/(dx) = 1/cos^2y (-1/x^2) = - 1/(x^2cos^2(1/x))#

Substitute this in the equation above and you have:

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

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

We can simplify this expression as:

#(dy)/(dx) = (2xsin(1/x)cos(1/x) - 1)/(cos^2(1/x)) = 2((xsin(2/x)-1)/(1+cos(2/x)))#