Question

What is the equation of the tangent line of `f(x)=2x^2+xe^x` at `x=5`?

Answer 1

Equation of tangent line is `(20+6e^5)x-y=50+30e^5`

Explanation:

The slope of the tangent to curve `y=f(x)` is given by `f'(x)`

Here `f(x)=2x^2+xe^x` and `f'(x)=4x+e^x+xe^x=4x+e^x(1+x)`

As we are seeking tangent at `x=5`,

we are seeking tangent at `(5,50+5e^5)`

and slope at `x=5` is `20+e^5*6=20+6e^5`

Hence equation of tangent is

`y-50-5e^5=(20+6e^5)(x-5)`

or `y-50-5e^5=20x-100+6e^5x-30e^5`

or `(20+6e^5)x-y=50+30e^5`

Answer 2

See below.

Explanation:

First we need to differentiate the function to find the gradient of the tangent line.

We need to use the product rule for the second term.

Product rule states that:

`f'(a*b)=b*f'(a)+a*f'(b)`

Let `a= x and b=e^x`

`f'(a*b)=e^x*1+x*e^x=e^x+xe^x`

`f'(2x^2)=4x`

`:.`

`f'(2x^2+xe^x)=4x+e^x+xe^x`

`f'(5)=4(5)+e^5+5e^5=20+6e^5`

Equation of line:

`y-(50+5e^5)=(20+6e^5)(x-5)`

`y=(20+6e^5)x-25(2+e^5)`