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Topics
Pre-Algebra
Mean
Mode
Greatest Common Factor
Least Common Multiple
Order of Operations
Fractions
Mixed Fractions
Prime Factorization
Exponents
Radicals
Algebra
Combine Like Terms
Solve for a Variable
Factor
Expand
Evaluate Fractions
Linear Equations
Quadratic Equations
Inequalities
Systems of Equations
Matrices
Trigonometry
Simplify
Evaluate
Graphs
Solve Equations
Calculus
Derivatives
Integrals
Limits
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Differentiation
10 problems similar to:
\frac { d } { d x } ( 2 )
$dxd (2)$
Similar Problems from Web Search
let f be a differentiable function. Compute \frac{d}{dx}g(2), where g(x) = \frac{f(2x)}{x}.
let
$f$
be a differentiable function. Compute
$dxd g(2)$
, where
$g(x)=xf(2x) $
.
https://math.stackexchange.com/questions/2351494/let-f-be-a-differentiable-function-compute-fracddxg2-where-gx
You have an extra 4 in the numerator here: i know that : \dfrac{d}{dx}g(2)=\dfrac{4(\dfrac{d}{dx}f(4))-4f(4)}{4} If g(x) = \dfrac{f(2x)}x, then \begin{align*} \frac d{dx} g(x) &= \frac d{dx} ...
You have an extra
$4$
in the numerator here: i know that :
$dxd g(2)=44(dxd f(4))−4f(4) $
If
$g(x)=xf(2x) $
, then \begin{align*} \frac d{dx} g(x) &= \frac d{dx} ...
How to rewrite \frac{d}{d(x+c)}? [closed]
How to rewrite
$d(x+c)d $
? [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} ...
Use the chain rule. Define
$u=x+c$
then use the fact that
$dxd⋅ =dxdu dud⋅ $
where the
$⋅$
represents any function, so
$dxdf =dxdu dudf $
...
What does is the meaning of \frac{d}{dx}+x in (\frac{d}{dx}+x)y=0?
What does is the meaning of
$dxd +x$
in
$(dxd +x)y=0$
?
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 ...
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?
Intuitive explanation of
$dxd =0$
?
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 ...
Not sure about the problem but the strength of the electrical field,
$E$
, depends on your distance from it, which I assume is
$x$
.
$dxdE $
then, is how much the strength of the field changes ...
Question about the chain rule.
Question about the chain rule.
https://math.stackexchange.com/q/2940216
Suppose we add an infinitesimal to x : x_1=x_0+\Delta x . What happens to y ? By definition, the derivative tells us how much a function changes relative to changes in its input: the change ...
Suppose we add an infinitesimal to
$x$
:
$x_{1}=x_{0}+Δx$
. What happens to
$y$
? By definition, the derivative tells us how much a function changes relative to changes in its input: the change ...
Spectrum of the derivative operator
Spectrum of the derivative operator
https://math.stackexchange.com/questions/2117107/spectrum-of-the-derivative-operator
\newcommand{\id}{I} As it was mentioned in the comments, the domain where you defined the operator is not correct - If you take C^1-functions with derivatives in L^2 the domain will be "too ...
As it was mentioned in the comments, the domain where you defined the operator is not correct - If you take
$C_{1}$
-functions with derivatives in
$L_{2}$
the domain will be "too ...
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Similar Problems
\frac { d } { d x } ( 2 )
$dxd (2)$
\frac { d } { d x } ( 4 x )
$dxd (4x)$
\frac { d } { d x } ( 6 x ^ 2 )
$dxd (6x_{2})$
\frac { d } { d x } ( 3x+7 )
$dxd (3x+7)$
\frac { d } { d a } ( 6a ( a -2) )
$dad (6a(a−2))$
\frac { d } { d z } ( \frac{z+3}{2z-4} )
$dzd (2z−4z+3 )$
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