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Algebra Baldor
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Find the derivative of $\tan\left(\arccos\left(x^2\right)\right)$

Used Formulas

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Solution

Formulas

Videos

Derivatives of trigonometric functions

· Derivative of tangent function
$\frac{d}{dx}\left(\tan\left(\theta \right)\right)=\frac{d}{dx}\left(\theta \right)\sec\left(\theta \right)^2$

Derivatives of inverse trigonometric functions

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

Basic Derivatives

· Power rule for derivatives
$\frac{d}{dx}\left(x^a\right)=ax^{\left(a-1\right)}$

Function Plot

Plotting: $\frac{-2x\sec\left(\arccos\left(x^2\right)\right)^2}{\sqrt{1-x^{4}}}$

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5
6
7
8
9
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

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

$\frac{d}{dx}\left(\arccot\left(\sqrt{x}\right)\right)$

$\frac{d}{dx}\left(\arcsin\sqrt{x}\right)$

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

$\frac{d}{dx}\arccos\left(\sqrt{x}\right)$

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

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

Main Topic: Differential Calculus

The derivative of a function of a real variable measures the sensitivity to change of a quantity (a function value or dependent variable) which is determined by another quantity (the independent variable). Derivatives are a fundamental tool of calculus.

Used Formulas

See formulas (3)

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