The derivative of root cosx is equal to -sinx/(2√cosx). In this post, we will learn how to differentiate the square root of cos x by the chain rule of derivatives. Derivative of Square Root of cos x by Chain Rule Question: Find the derivative of $\sqrt{\cos x}$ by the chain rule. Solution: To find the … Read more
The derivative of root sinx is equal to cosx/(2√sinx). In this post, we will learn how to differentiate square root of sinx by the chain rule of derivatives. Derivative of Square Root of Sin x by Chain Rule Question: Find the Derivative of $\sqrt{\sin x}$ by the chain rule. Solution: To find the derivative of … Read more
The limit of (ex-1)/x, when x tends to zero, is equal to 1. That is, the formula of the limx→0 (ex-1)/x is given by limx→0 (ex-1)/x = 1. In this post, we will learn to find the limit of the function (ex-1)/x when x approaches to zero. Limit of (e^x-1)/x as x→0 Prove that limx→0 … Read more
The limit of sinx/x as x approaches infinity is equal to zero. That is, limx→∞ sinx/x = 0. This can be proved using the Squeeze Theorem of limits which will be done here. Limit of sinx/x as x goes to infinity Question: What is the limit of sinx/x when x→∞. Answer: The limit of sinx/x … Read more
The derivative of 1/cos is equal to secx tanx. In this post, we will learn how to differentiate 1 over cosx with respect to x. What is the Derivative of 1/cosx? Answer: The derivative of 1/cosx with respect to x is denoted by d/dx(1/cosx) and it is given below. $\dfrac{d}{dx}(\dfrac{1}{\cos x})$ = $\sec x \tan … Read more
The derivative of 1/sinx is equal to -cosecx cotx. In this post, we will learn how to differentiate 1 by sinx with respect to x. What is the Derivative of 1/sinx? Answer: The derivative of 1/sinx with respect to x is denoted by d/dx(1/sinx) and it is equal to -cosec(x)cot(x). That is, $\dfrac{d}{dx}(\dfrac{1}{\sin x})$ = … Read more
The Maclaurin series expansion of ex or the Taylor series expansion of ex at x=0 is given by the following summation: ex = $\sum_{n=0}^\infty \dfrac{x^n}{n!}$ = $1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\cdots$. In this post, we will learn how to find the series expansion of ex. Taylor Series Expansion of ex at x=0 The Maclaurin series expansion of a function … Read more
For any real number c, we have limx→ccosx = cosc and cosx is defined for all real numbers. Thus, the function cosx is continuous everywhere. Now, we will prove that cosx is continuous for all values of x by the epsilon-delta method. We will use the following two formulas: Prove cosx is continuous Let f(x)=cosx … Read more
The Maclaurin series expansion of cosx or the Taylor series expansion of cosx at x=0 is given as follows: cosx = $\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!}x^{2n}$ = $1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\cdots$ Taylor Series Expansion of Sinx at x=0 Note that the Maclaurin series expansion of f(x)=cosx or the Taylor series of a function f(x) at $x=0$ is given by the following … Read more
The value of the limit of x/sinx as x approaches 0 is equal to 1. In this post, we will learn how to find the limit of $\frac{x}{\sin x}$ when x tends to 0. lim x/sinx when x approaches 0 Formula The formula of the limit of x/sinx when x tends to zero is given … Read more