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Vejledning i brug af definition af en afledning
\frac { d } { d x } ( 4 x )
Afledningen af ax^{n} er nax^{n-1}.
4x^{1-1}
Subtraher 1 fra 1.
4x^{0}
For ethvert led t bortset fra 0, t^{0}=1.
4\times 1
For ethvert led t, t\times 1=t og 1t=t.
4
Differentier w.r.t. x
0
Quiz
Differentiation
\frac { d } { d x } ( 4 x )
Lignende problemer fra websøgning
How to calculate \frac {\mathrm d}{\mathrm dx} {x!} ?
https://math.stackexchange.com/questions/2097127/how-to-calculate-frac-mathrm-d-mathrm-dx-x
It does not a priori make sense to differentiate x! because the domain of x\mapsto x! is \mathbf N, not \mathbf R (or anything else supporting a good notion of differentiation, like \mathbf C ...
How to rewrite \frac{d}{d(x+c)}? [closed]
https://math.stackexchange.com/questions/1376627/how-to-rewrite-fracddxc
Use the chain rule. Define u = x + c then use the fact that \frac{d\cdot}{dx} = \frac{du}{dx} \frac{d\cdot}{du} where the \cdot represents any function, so \frac{df}{dx} = \frac{du}{dx} \frac{df}{du} ...
What does is the meaning of \frac{d}{dx}+x in (\frac{d}{dx}+x)y=0?
https://math.stackexchange.com/q/1590756
The symbols d/dx and x should both be interpreted as linear operators acting on a vector space that the unknown function y belongs to. The sum of linear operators is well-defined and that is ...
Intuitive explanation of \frac{\mathrm{d}}{\mathrm{d}x}=0?
https://math.stackexchange.com/questions/2894024/intuitive-explanation-of-frac-mathrmd-mathrmdx-0
Not sure about the problem but the strength of the electrical field, E, depends on your distance from it, which I assume is x. \frac{dE}{dx} then, is how much the strength of the field changes ...
Differentiating the polynomial x^3 - 4x +6
https://math.stackexchange.com/q/65332
Everything is correct, except that the derivative of a constant (like 6) is always 0. You can still see this fact from the power rule. Write 6 as 6x^0. The power rule says that the derivative is 6 \cdot 0 x^{-1} ...
How do I handle dx in u-substitution?
https://math.stackexchange.com/questions/927265/how-do-i-handle-dx-in-u-substitution
If you choose u=x^2+1, then taking the derivative with respect to x gives: \frac{\textrm{d}u}{\textrm{d}x}=2x. Therefore \textrm{d}u=2x\textrm{d}x or x\textrm{d}x=\textrm{d}u/2. Now your ...
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4x^{1-1}
Afledningen af ax^{n} er nax^{n-1}.
4x^{0}
Subtraher 1 fra 1.
4\times 1
For ethvert led t bortset fra 0, t^{0}=1.
4
For ethvert led t, t\times 1=t og 1t=t.
Lignende problemer
\frac { d } { d x } ( 2 )
\frac { d } { d x } ( 4 x )
\frac { d } { d x } ( 6 x ^ 2 )
\frac { d } { d x } ( 3x+7 )
\frac { d } { d a } ( 6a ( a -2) )
\frac { d } { d z } ( \frac{z+3}{2z-4} )
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