What is the equation of the tangent line of #f(x)=(x^2)e^(x+2) # at #x=2#?

1 Answer
Oct 22, 2016

Therefore,the equation of the tangent line is #color(purple)(y=8e^4x-12e^4)#

Explanation:

The tangent line passes through the point #(2,f(2))#
#color(purple)(f(2))=2^2*e^(2+2)=4e^4#

To write its equation we should find its slope#color(red)(f'(2)#

Applying the following properties on derivative we can compute #color(red)(f'(x))#

Derivative of products:
#color(blue)((duv)/dx=u'v+v'u)#

Derivatives on power:
#color(orange)((dx^n)/dx=bx ^(n-1)#

Derivatives of exponential function :
#color(brown)((de^(u(x)))/dx=(du(x))/dx*e^(u(x)))#

#f(x)=(x^2)e^(x+2)#
#color(red)(f'(x))=color(blue)((dx^2)/dx*e^(x+2)+(de^(x+2))/dx*x^2)#
#color(red)(f'(x))=color(orange)(2x)e^(x+2)+color(brown)(e^(x+2))*x^2#
#color(red)(f'(x)=xe^(x+2)(2+x))#

#color(red)(f'(2))=2e^4*4=8e^4#

So,the equation of the tangent line passing through#(2,4e^4)# and of slope #8e^4# is:
#y-color(purple)(f(2))=color(red)(f'(2))(x-2)#
#y-4e^4=8e^4(x-2)#
#y-4e^4=8e^4x-16e^4#
#y=8e^4x-16e^4+4e^4#
#y=8e^4x-12e^4#

Therefore,the equation of the tangent line is #color(purple)(y=8e^4x-12e^4)#