Question

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

Answer 1

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)`