What is the equation of the line tangent to # f(x)=x^2e^x-xe^(x^2) # at # x=0#?

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Answer 1 · 2017-12-19T07:03:37

Let's see.

Given,

#f(x)=x^2e^x+xe^(x^2)#

Differentiate it and find the slope #dy/dx#.

Now find the #dy/dx(x=0)#

This slope is the slope of the tangent at that coordinate.

Now find the #f(0)#.

#:.x_1=0, y_1=f(0)#

#:.x_1=0, y_1=0#

Now use the equation of straight lines #rarr#

#y-y_1=dy/dx(x-x_1)#

#:.y-0=dy/dx(x=0)cdot(x-0)#

#:.y/x=dy/dx(x=0)#

Substitute the value of #dy/dx(x=0)# and get the suitable equation of the tangent.

Hope it Helps:)