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Find the derivative of $\frac{1}{2}\tan\left(x\right)\sin\left(2x\right)$

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Solution

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Videos

Derivatives of sec(x) and csc(x) | Derivative rules | AP Calculus AB | Khan Academy

https://www.youtube.com/watch?v=TDJ5nXWEkWM

Calculus: Derivatives 2 | Taking derivatives | Differential Calculus | Khan Academy

https://www.youtube.com/watch?v=ay8838UZ4nM

Calculus - Take the log of both sides to find the derivative, y = (x(x^2 + 1)^2)/(sqrt(2x^2 - 1))

https://www.youtube.com/watch?v=7aF6Ck6ZRxw

Implicit differentiation | Advanced derivatives | AP Calculus AB | Khan Academy

https://www.youtube.com/watch?v=mSVrqKZDRF4

Calculus - How find the first derivative using a chart, 1/f(x) at x=1

https://www.youtube.com/watch?v=V5OROgIBnDc

Linear approximation of a rational function | Derivative rules | AP Calculus AB | Khan Academy

https://www.youtube.com/watch?v=vRw4ovZK1CQ

Function Plot

Plotting: $\frac{1}{2}\sec\left(x\right)^2\sin\left(2x\right)+\tan\left(x\right)\cos\left(2x\right)$

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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:

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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 (6)

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