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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
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Systems of Equations
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Trigonometry
Simplify
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4
$4$
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Steps Using Definition of a Derivative
\frac { d } { d x } ( 4 x )
$dxd (4x)$
The derivative of ax^{n} is nax^{n-1}.
The derivative of
$ax_{n}$
is
$nax_{n−1}$
.
4x^{1-1}
$4x_{1−1}$
Subtract 1 from 1.
Subtract
$1$
from
$1$
.
4x^{0}
$4x_{0}$
For any term t except 0, t^{0}=1.
For any term
$t$
except
$0$
,
$t_{0}=1$
.
4\times 1
$4×1$
For any term t, t\times 1=t and 1t=t.
For any term
$t$
,
$t×1=t$
and
$1t=t$
.
4
$4$
Differentiate w.r.t. x
0
$0$
Graph
Quiz
Differentiation
5 problems similar to:
\frac { d } { d x } ( 4 x )
$dxd (4x)$
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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 ...
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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
$u=x+c$
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$dxd⋅ =dxdu dud⋅ $
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$⋅$
represents any function, so
$dxdf =dxdu dudf $
...
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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 ...
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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 ...
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Differentiating the polynomial x^3 - 4x +6
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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} ...
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
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How do I handle dx in u-substitution?
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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 ...
If you choose
$u=x_{2}+1$
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4x^{1-1}
The derivative of ax^{n} is nax^{n-1}.
4x^{0}
Subtract 1 from 1.
4\times 1
For any term t except 0, t^{0}=1.
4
For any term t, t\times 1=t and 1t=t.
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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