Februar 9, 2020

von Bianca

Wie unterscheidet man $y = ln (1 + x^{2})$ $y = ln (1 + x ^ 2)$ ?

Antworten:

$\frac{d y}{d x} = \frac{2 x}{1 + x^{2}}$ $dy/dx=(2x)/(1+x^2)$

Erläuterung:

differentiate using the $chain rule$ $color(blue)"chain rule"$

That is $∣ ∣ ∣ ∣ ∣ ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ \frac{a}{a} \frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x} \frac{a}{a} ∣ ∣ ∣ - ----------------- - ... ... . . (A)$ $color(red)(|bar(ul(color(white)(a/a)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(a/a)|)))........ (A)$

$Reminder ∣ ∣ ∣ ∣ ∣ ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ \frac{a}{a} \frac{d}{d x} (ln x) = \frac{1}{x} \frac{a}{a} ∣ ∣ ∣ - --------------- -$ $color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(lnx)=1/x)color(white)(a/a)|)))$

let $u = 1 + x^{2} \Rightarrow \frac{d u}{d x} = 2 x$ $u=1+x^2rArr(du)/(dx)=2x$

and so $y = ln u \Rightarrow \frac{d y}{d u} = \frac{1}{u}$ $y=lnurArr(dy)/(du)=1/u$

substitute these values into (A) changing u back to terms of x.

$\Rightarrow \frac{d y}{d x} = \frac{1}{u} (2 x) = \frac{2 x}{1 + x^{2}}$ $rArrdy/dx=1/u(2x)=(2x)/(1+x^2)$