FPM(2023) January Paper 01R Q6 and solution

The curve $C$ has the equation:

$ y = \frac{e^{x^2+1}}{x^2 + 1}. $

(a) Show that

$ \frac{dy}{dx} = \frac{Kx^3 e^{x^2+1}}{(x^2 + 1)^2}, $

where $K$ is a constant whose value is to be found.

Solution:

Let $y = \frac{u}{v}$, where $u = e^{x^2+1}$ and $v = x^2 + 1$. Using the quotient rule:

$ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}. $

First, find $\frac{du}{dx}$:

$ u = e^{x^2+1}, \quad \frac{du}{dx} = e^{x^2+1} \cdot \frac{d}{dx}(x^2 + 1) = e^{x^2+1} \cdot 2x. $

Next, find $\frac{dv}{dx}$:

$ v = x^2 + 1, \quad \frac{dv}{dx} = 2x. $

Substitute into the quotient rule:

$ \frac{dy}{dx} = \frac{(x^2 + 1)(2x e^{x^2+1}) - e^{x^2+1}(2x)}{(x^2 + 1)^2}. $

Factorize the numerator:

$ \frac{dy}{dx} = \frac{2x e^{x^2+1} \left((x^2 + 1) - 1\right)}{(x^2 + 1)^2}. $

$ \frac{dy}{dx} = \frac{2x e^{x^2+1} \cdot x^2}{(x^2 + 1)^2}. $

Simplify:

$ \frac{dy}{dx} = \frac{2x^3 e^{x^2+1}}{(x^2 + 1)^2}. $

Thus, $K = 2$.

(b) Find an equation of the tangent to $C$ at $x = -1$

Substitute $x = -1$ into $y = \frac{e^{x^2+1}}{x^2 + 1}$ to find the point of tangency:

$ y = \frac{e^{(-1)^2+1}}{(-1)^2 + 1} = \frac{e^{2}}{2}. $

So, the point is:

$ \left(-1, \frac{e^2}{2}\right). $

Next, find the slope of the tangent at $x = -1$ using $\frac{dy}{dx}$:

$ \frac{dy}{dx} = \frac{2x^3 e^{x^2+1}}{(x^2 + 1)^2}. $

Substitute $x = -1$:

$ \frac{dy}{dx} = \frac{2(-1)^3 e^{(-1)^2+1}}{((-1)^2 + 1)^2}. $

$ \frac{dy}{dx} = \frac{2(-1)^3 e^2}{(1 + 1)^2}. $

$ \frac{dy}{dx} = \frac{-2 e^2}{4} = -\frac{e^2}{2}. $

The equation of the tangent is:

$ y - y_1 = m(x - x_1), $

where $\left(x_1, y_1\right) = \left(-1, \frac{e^2}{2}\right)$ and $m = -\frac{e^2}{2}$. Substitute:

$ y - \frac{e^2}{2} = -\frac{e^2}{2}(x + 1). $

Simplify:

$ y = -\frac{e^2}{2}x - \frac{e^2}{2} + \frac{e^2}{2}. $

$ y = -\frac{e^2}{2}x. $

Final Answers:

• (a) $K = 2$.
• (b) The equation of the tangent is:

$ y = -\frac{e^2}{2}x. $