How do you find the equation of the tangent line to the graph #y=xe^x-e^x# through point (1,0)?

1 Answer
Andrew B.
Mar 3, 2017

#y=xe-e#

Explanation:

Recall that the product rule states that for the function #f(x)=g(x)*h(x)#

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

For this function, if #y=f(x)=xe^x-e^x#

Factor out #e^x#
#f(x)=e^x(x-1)#

Apply the product rule to differentiate
#f'(x)=e^x(x-1)+e^x(1)#

Simplify
#f'(x)=e^x(x-1+1)=xe^x#

Recall that the derivative of a function at a point is equal to the rate of change of that function at that point. In other words, the derivative of #f(x)# at #(1,0)# is equal to the slope of the tangent line at #(1,0)#

Since we have found the derivative function of #f(x)#, plug in 1 to #f'(x)# to get the slope of the tangent line.

#f'(1)=(1)e^1=e#

Now, we have the slope of a line and one point that it goes through. We can apply the point-slope form of a line to find its equation.

#y-y_1=m(x-x_1)#

where:
#y_1# is #f(1)#, #m=f'(1)#, and #x_1=1#

Subbing in, we get:

#y-y_1=m(x-x_1)#

#y-0=e(x-1)#

#y=xe-e#

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