Question #7055d

Question

Question #7055d

2 Answers

Impact of this question

2256 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-29T19:15:00

Make sure that you know your exponent rules!

Explanation:

Let's quickly look at the product rule for exponents.

$a^{n} \cdot a^{m} = a^{n + n}$ $color(green)(a^n * a^m = a^(n + n)$

You thought the power rule for exponents was

$a^{n} \cdot a^{m} = a^{n \cdot m}$ $color(red)(a^n * a^m = a^(n * m)$ , which is wrong

Our only choice to differentiate $e^{x^{2}}$ $e^(x^2)$ would be to use the chain rule. Let $y = e^{u}$ $y= e^u$ and $u = x^{2}$ $u = x^2$ . Then $\frac{d y}{d u} = e^{u}$ $dy/(du) = e^u$ and $\frac{d u}{d x} = 2 x$ $(du)/dx = 2x$ .

The chain rule states that $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $color(magenta)(dy/dx= dy/(du) * (du)/dx$ . This is obviously true since when multiplied, the $d u$ $du$ 's cancel to leave $\frac{d y}{d x}$ $dy/dx$ on both sides.

$\frac{d y}{d x} = e^{u} \cdot 2 x$ $dy/dx = e^u * 2x$

$\frac{d y}{d x} = 2 x e^{x^{2}} \to$ $dy/dx= 2xe^(x^2)->$ This is the correct derivative

Hopefully this helps!

Answer 2 · 2017-01-29T19:16:01

$2 x e^{x^{2}} and 2 e^{2 x}$ $2xe^(x^2)" and " 2e^(2x)$

Explanation:

Note that $e^{x^{2}} \neq e^{x} \times e^{x} but e^{x} \times e^{x} = e^{2 x}$ $e^(x^2)≠e^x xxe^x" but "e^x xxe^x=e^(2x)$

Using the $standard derivative of the exponential function$ $color(blue)"standard derivative of the exponential function"$

$color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx(e^(f(x)))=e^(f(x)).f'(x))color(white)(2/2)|)))$
$color(white)(xxxxxxxx)"A version of the " color(blue)"chain rule"$

$rArrd/dx(e^(x^2))=e^(x^2).d/dx(x^2)=2xe^(x^2)$

$"and "d/dx(e^(2x))=e^(2x).d/dx(2x)=2e^(2x)$

Science

Math

Humanities

... and beyond

Question #7055d

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question