How do you differentiate $y = e^{x^{2}}$ $y=e^(x^2)$ ?

Question

How do you differentiate $y = e^{x^{2}}$ $y=e^(x^2)$ ?

1 Answer

Impact of this question

77962 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-17T08:52:53

$\frac{d y}{d x} = 2 x e^{x^{2}}$ $(dy)/(dx)=2xe^(x^2)$

Explanation:

Chain Rule - In order to differentiate a function of a function, say $y, = f (g (x))$ $y, =f(g(x))$ , where we have to find $\frac{d y}{d x}$ $(dy)/(dx)$ , we need to do (a) substitute $u = g (x)$ $u=g(x)$ , which gives us $y = f (u)$ $y=f(u)$ . Then we need to use a formula called Chain Rule, which states that $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)xx(du)/(dx)$ . In fact if we have something like $y = f (g (h (x)))$ $y=f(g(h(x)))$ , we can have $\frac{d y}{d x} = \frac{d y}{d f} \times \frac{d f}{d g} \times \frac{d g}{d h}$ $(dy)/(dx)=(dy)/(df)xx(df)/(dg)xx(dg)/(dh)$

Here we have $y = e^{u}$ $y=e^u$ , where $u = x^{2}$ $u=x^2$

Hence, $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)xx(du)/(dx)$

= $\frac{d}{d u} e^{u} \times \frac{d}{d x} (x^{2})$ $d/(du)e^uxxd/dx(x^2)$

= $e^{u} \times 2 x = 2 x e^{x^{2}}$ $e^uxx2x=2xe^(x^2)$

Science

Math

Humanities

... and beyond

How do you differentiate $y = e^{x^{2}}$ $y=e^(x^2)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you differentiate y=ex2y=e^(x^2)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you differentiate $y = e^{x^{2}}$ $y=e^(x^2)$ ?