Question

LINEAR APPROXIMATIONS AND DIFFERENTIALS;
"Find the differential of each function."
y=[sin(x)]/x
y=ln[4+x^(2)]^(2)

I am aware that the formula for differentials is Δy=f(x+Δx)-f(x)

Answer 1

LaTeX form:\[y=\frac{\sin{(x)}}{x}\]\[y=\ln{(4+x^{2})}^{2}\]

Answer 2

Given \(y=f(x)\), the differential would be \(\mathrm dy=\dfrac{\mathrm df(x)}{\mathrm dx}\,\mathrm dx\). The formula you refer to is an approximation of this.

Answer 3

Oooh, okay. That makes more sense.
Sorry, my brain is kind of out of it today ahah

Answer 4

... er, how would I write \(df(x)\) for an equation?

Answer 5

Well, \(\mathrm d(\text{something})\) is the differential for \(\text{something}\), while \(\dfrac{\mathrm d(\text{something})}{\mathrm dx}\) is the derivative of \(\text{something}\) with respect to the variable \(x\). The derivative doesn't technically work like a ratio, so the \(\mathrm dx\)s don't cancel out (though in some contexts it's useful and sometimes intuitive to act like they do).

So when I say the differential of \(y=f(x)\) is \(\mathrm dy=\dfrac{\mathrm df(x)}{\mathrm dx}\,\mathrm dx\), all this means is that the differential of \(y\) (the dependent variable) is the product of the derivative of \(y\) with respect to \(x\) and the differential of \(x\) (the independent variable).

For example, if \(y=x^2\), then \(\mathrm dy=\dfrac{\mathrm d}{\mathrm dx}[x^2]\,\mathrm dx=2x\,\mathrm dx\).

Answer 6

Maybe it helps to write \(\dfrac{\mathrm df(x)}{\mathrm dx}\) as \(\dfrac{\mathrm d}{\mathrm dx}[f(x)]\) ? I think the latter notation is a bit clearer in indicating that \(\dfrac{\mathrm d}{\mathrm dx}\) is an operator as opposed to a ratio of differentials. Both mean the same thing, and that is the derivative \(f'(x)\). Calculus notation can be very confusing.

Answer 7

Alright... I think I get it.
My main issue is applying the formulaic properties to the equations, really. I somewhat get the concepts aha

Answer 8

take a look at the formula u are given

Answer 9

@HolsterEmission is wrt \(x\) derivative implicit?

Answer 10

The formula you posted can be derived geometrically.

The slope of this tangent line is the ratio of the change in \(y\) to the change in \(x\):
\[\frac{\Delta y}{\Delta x}=\frac{f(b)-f(a)}{b-a}\]Since \(b-a=\Delta x\), you can write \(b=a+\Delta x\), so the slope is
\[\frac{\Delta y}{\Delta x}=\frac{f(a+\Delta x)-f(a)}{\Delta x}\]and as long as \(a\neq b\), then \(\Delta x\neq0\), so you can multiply both sides by this number to get the approximating differential formula you're given:
\[\Delta y=\frac{f(a+\Delta x)-f(a)}{\Delta x}\Delta x=f(a+\Delta x)-f(a)\](and you can replace \(a\) with any symbol you like).

(Second comment incoming, just need to publish this first drawing before I post it...)