A curve has the equation $y = a^{x^2}$ , where $a\,$ is a constant greater than 1.

Show that $dydx=2xax2ln⁡a\displaystyle \frac{dy}{dx} = 2x a^{x^2} \ln a$

The tangent at the point $2, a^4)$ passes through the point $(1, 0)$ . Find the value of $a$ , giving your answer in an exact form.

By considering $d2ydx2\displaystyle \frac{d^2y}{dx^2}$ show that the curve is convex for all values of $x$ .

A curve has the equation $y = a^{x^2}$ , where $a\,$ is a constant greater than 1.

Show that $dydx=2xax2ln⁡a\displaystyle \frac{dy}{dx} = 2x a^{x^2} \ln a$

The tangent at the point $2, a^4)$ passes through the point $(1, 0)$ . Find the value of $a$ , giving your answer in an exact form.

By considering $d2ydx2\displaystyle \frac{d^2y}{dx^2}$ show that the curve is convex for all values of $x$ .

Differentiation

Differentiation Questions