What is the equation of the line tangent to #f(x)= xe^x # at #x=7#?

Let #f(x) = g(x) xx h(x)#, then #g(x) = x# and #h(x) = e^x#. The product rule states that #f'(x) = g'(x) xx h(x) + g(x) xx h'(x)#

The derivative of #x# is #1#. The derivative of #e^x# is #e^x#.

Hence, #f'(x) = 1 xx e^x + xe^x =e^x + xe^x = e^x(1 + x)#

The function and the tangent pass through the point #(7, 7e^7)#.

All that we have left to find is the slope of the tangent. The slope of the tangent is given by evaluating #f'(a)#, where #x = a# is the given point.

#m_"tangent" = e^7(1 + 7) = 8e^7#

Now, by point-slope form we have:

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

#y - 7e^7 = 8e^7(x - 7)#

#y - 7e^7 = 8e^7x - 56e^7#

#y = 8e^7x - 49e^7#

In summary, the equation of the tangent of #f(x) = xe^x# at the point #x = 7# is #y = 8e^7x - 49e^7#.

Hopefully this helps!