Derivative of Cot x – Formula, Proof by First Principle, Chain & Quotient Rule
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One of the first transcendental functions introduced in Differential Calculus is the Derivative of Cotangent (or Calculus I).
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Let us go over the derivative of cot x formula, as well as its proof (in various methods) and a few solved examples.
What is Derivative Cotx?
Cot(x) is a trigonometric function that represents the ratio of the adjacent side to the opposite side of a right triangle where the angle opposite the adjacent side is x. More specifically, cot(x) is defined as the reciprocal of the tangent function, or cot(x) = 1/tan(x).
Alternatively, it can be expressed in terms of the cosine and sine functions as cot(x) = cos(x)/sin(x).
The cotangent function is periodic with a period of π, and it has singularities at the zeros of the sine function, which correspond to the vertical asymptotes of the graph. The range of cot(x) is all real numbers, except for 0, which is not in the range since sin(x) and cos(x) cannot be zero simultaneously.
Derivative of cot x
If x is used to represent a variable, the cotangent function is written as cot x in mathematics.
The differentiation of the cot function with respect to x is written in differential calculus in the following mathematical form.
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Derivative of Cotx Formula
The derivative of cot(x) is given by the formula d/dx(cot(x)) = -csc^2(x), where "csc" stands for cosecant. This formula represents the rate of change of the cotangent function with respect to its input x. It states that the slope of cot(x) at any given point x is equal to the negative cosecant squared of that point.
The derivative of cot(x) is an important concept in calculus, and it is used to solve various problems involving the rate of change, optimization, and integration.
Derivative of Cot x Proof by First Principle
We assume that
Step 1:
And,
Step 2:
Step 3:
In equation (1) subtract these values
Step 4:
Step 5:
Step 6:
Using the sum and difference formulas,
Step 7:
Step 8:
Step 9:
We have,
Step 10:
Step 11:
Step 12:
Step 13:
We know that sin reciprocal is csc. So,
Hence proved
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Derivative of Cot x Proof by Chain Rule
We can use the chain rule to prove the derivative of the cot x formula. Let us remember that cot and tan are reciprocals of each other.
The power rule can be applied here. According to the power and chain rules,
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Step 6:
Step 7:
We know that sin’s reciprocal is csc.
Hence proved
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Derivative of Cot x Proof by Quotient Rule
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Step 6:
Step 7:
Step 8:
Hence proved
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Common Misconceptions Related to Derivative of Cot x:
Here are some common misunderstandings people have when finding the derivative of cot x. Let’s clear them up:
Though cot x = (cos x)/(sin x), the derivative of cot x is NOT the same as the derivative of cos x divided by the derivative of sin x.
We must apply the quotient rule to find the derivative of cot x (since it's written as a division of cos x by sin x).
The derivative of cot x is NOT equal to tan x. Remember, cot x and tan x are reciprocals, but that doesn’t mean their derivatives are also reciprocals.
Also, the derivative of cot x is not the same as the derivative of cot inverse x.
Solved Examples Using Derivative of cot x
Problem 1:Find the derivative of f(x)=cotx
Solution:
We know that the derivative of cot x is:
d/dx(cotx)=−csc2
So,
f′(x)=−csc2x
Problem 2: Find the derivative of 1/cotx
Solution:
We rewrite 1/cot x as tan x because cot x = 1/tan x. Therefore, 1/cot x = tan x.
Now, we can use the derivative formula for tan x which is sec^2 x. Therefore:
So the derivative of 1/cot x with respect to x is sec^2 x.
Problem 3: Find the derivative of cotx sinx
Solution: We can start by using the product rule of differentiation, which states that for two functions $u(x)$ and $v(x)$, the derivative of their product $u(x)v(x)$ is given by:
(u(x) v(x))′=u′(x) v(x) + u(x) v′(x)
(cotx sinx)′= (cotx)′ Sinx + cotx (sinx)′
Applying this rule to cot x sin x, we get:
We can simplify this expression by finding the derivatives of $\cot x$ and $\sin x$. The derivative of $\cot x$ is $-\csc^2 x$, and the derivative of $\sin x$ is $\cos x$. Therefore:
\(\begin{align*}(\cot x \sin x)' &= (-\csc^2 x)\sin x + \cot x (\cos x) \&= -\frac{\sin x}{\sin^2 x} + \frac{\cos x}{\sin x} \&= -\frac{1}{\sin x} + \frac{\cos x}{\sin x}
\end{align*}\)
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FAQs For Derivative of Cot X
What is cot x equal to?
What is the second derivative of cot X?
What is the integral of cot X?
the integral of cot X
Is the derivative of cot x equal to the derivative of cot inverse x?
Yes the derivative of cot x equal to the derivative of cot inverse x
What is the difference between the derivative of cot x and the antiderivative of cot x?
The derivative of cot x is
What is the domain of the derivative of cot(x)?
The derivative -csc²(x) is undefined where cot(x) is undefined, i.e., at x=nπ, where nnn is an integer.
Is the derivative of cot(x) positive or negative?
The derivative of cot(x) is negative, as it is -csc²(x), and csc²(x) is always positive (where defined).