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Algebra Baldor
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Find the derivative of $\arcsin\left(2x-5\right)$

Used Formulas

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Solution

Formulas

Videos

Derivatives of inverse trigonometric functions

$\frac{d}{dx}\left(\arcsin\left(\theta \right)\right)=\frac{1}{\sqrt{1-\theta ^2}}\frac{d}{dx}\left(\theta \right)$

Basic Differentiation Rules

· Derivative of a Constant
$\frac{d}{dx}\left(c\right)=0$
· Sum Rule for Differentiation
$\frac{d}{dx}\left[f\left(x\right)+g\left(x\right)\right]=\frac{d}{dx}f\left(x\right) + \frac{d}{dx}g\left(x\right)$
$\frac{d}{dx}\left(cx\right)=c\frac{d}{dx}\left(x\right)$
· Derivative of the linear function
$\frac{d}{dx}\left(x\right)=1$

Function Plot

Plotting: $\frac{2}{\sqrt{1-\left(2x-5\right)^2}}$

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0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Similar Problems Solved

$\frac{d}{dx}\arctan\left(x-\sqrt{1+x^2}\right)$

$\frac{d}{dx}\arccos\left(\sqrt{1-x^2}\right)$

$\frac{d}{dx}\left(\arcsin\left(1-2x^2\right)\right)$

$\frac{d}{dx}\arcsin\left(2x-3\right)$

$\frac{d}{dx}\left(\arcsin\left(2x-5\right)\right)$

$\frac{d}{dx}\left(x\sqrt{x^2-1}\right)$

$\frac{d}{dx}\left(2x+1\right)^3$

Main Topic: Sum Rule of Differentiation

The sum rule is a method to find the derivative of a function that is the sum of two or more functions.

Used Formulas

See formulas (5)

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