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Evaluating Limits : Evaluating Limits: A Step-by-Step Guide : Learn the easy method of Evaluating Limits by direct substitution. Master calculus with our step-by-step guide and examples!

Evaluating Limits

Learn how to easily evaluate limits using direct substitution! This guide provides clear examples and explanations to help you master this essential calculus skill. The SEO Keyphrase is Evaluating Limits. READ MORE...

understanding derivatives simply : Understanding Derivatives Simply: A Clear Explanation : Need to understand derivatives simply? This beginner's guide breaks down the concept with clear explanations and examples.

Understanding Derivatives Simply: A Beginner's Guide

Understanding derivatives simply involves grasping how functions change. This guide offers an intuitive explanation for beginners. READ MORE...

fractional differentiability functions : Fractional Differentiability Functions: A Deep Dive : Understand fractional differentiability functions, power functions, and absolute values. Construct functions with specific differentiability properties.

Fractional Differentiability of Functions

Explore fractional differentiability functions and how they behave with non-integer derivatives. Learn about constructing functions with specific differentiability. READ MORE...

minimum value function : Minimum Value Function: Calculus Guide and Examples : Discover how to find the minimum value function using partial derivatives and the second derivative test. Includes detailed examples.

Finding Minimum Value Function

Learn how to find the minimum value function using calculus. Step-by-step guide included! READ MORE...

Derivative of \(\mathsf { x^{n} }\) using the First Principle

Let y = \(\mathsf {x^{n} }\) โˆด y + ฮดy = \(\mathsf { {(x + ฮดx)^{n}} }\) โˆด ฮดy = y + ฮดy - y = \(\mathsf { (x + ฮดx)^{n} }\) - \(\mathsf { x^{n} }\) or ฮดy = \(\mathsf { [\text{ }^{n}C_0 x^{n}{(ฮดx)}^{0} }\) + \(\mathsf {\text{ }^{n}C_1 x^{n-1}{(ฮดx)}^{1}}\) + \(\mathsf {\text{ }^{n}C_2 x^{n-2}{(ฮดx)}^{2}}\) + \(\mathsf {\text{ }^{n}C_3 x^{n-3}{(ฮดx)}^{3}}\) \(\mathsf {+\ ... higher\ powers\ of\ ฮดx\ ] }\) - \(\mathsf {x^{n} }\) or ฮดy = \(\mathsf { [\text{ }^{n}C_0 x^{n} }\) + \(\mathsf {\text{ }^{n}C_1 x^{n-1}{(ฮดx)}^{1}}\) + \(\mathsf {\text{ }^{n}C_2 x^{n-2}{(ฮดx)}^{2}}\) + \(\mathsf {\text{ }^{n}C_3 x^{n-3}{(ฮดx)}^{3}}\) \(\mathsf{+\ ... \ ]} \) - \(\mathsf {x^{n} }\) Now, \(\mathsf { ^{n}C_0 x^{n} = 1.x^{n} = x^{n} }\) Cancelling \(\mathsf { […] READ MORE...

Derivative of \({e}^x\) using First Principleย 

Derivative of \({e}^x\) using the First Principle Let \(y\) = \({e}^x\)โˆด \(y + ฮดy\) = \({e}^{x + ฮดx}\)โˆด \(ฮดy\) = \({e}^{x + ฮดx}\) - \({e}^x\)or \(ฮดy\) = \({e}^{x}\) . \( [ {e}^{ฮดx} - 1 ]\)Dividing each side by ฮดx </h3>or \(\dfrac {ฮดy}{ฮดx}\) = \( \dfrac { {e}^{x} . [ {e}^{ฮดx} - 1 ] } {ฮดx}\) โˆด \(\dfrac {dy}{dx} = \) \( \lim_{ฮดx \to 0} \) \( \dfrac { {e}^{x} . [ {e}^{ฮดx} - 1 ] } {ฮดx}\)or \(\dfrac {dy}{dx}\) = \( {e}^{x} .\) \( \lim_{ฮดx \to 0} \) \( \dfrac { [ {e}^{ฮดx} - 1 ] } {ฮดx}\) ----- (1) Now, \( \lim_{ฮดx \to 0} \) \( \dfrac { [ {e}^{ฮดx} - 1 ] } {ฮดx}\) =\( \lim_{ฮดx \to 0} […] READ MORE...

Derivative of sinฮธ using the First Principle

Derivative of \( sinฮธ \) using the First Principle Let \(y\) = \( sinฮธ \) โˆด \(y + ฮดy\) = \( sin(ฮธ + ฮดฮธ) \) โˆด \(ฮดy\) = \( sin(ฮธ + ฮดฮธ) \) - \( sinฮธ \)From Trigonometry , we have \( sin(A-B) \) = 2.\( sin \dfrac {(A-B)}{2} \).\( cos \dfrac {(A+B)}{2} \)Using the above rule, we get\(ฮดy\) = 2.\( sin \dfrac {(ฮธ+ฮดฮธ - ฮธ)}{2} \).\( cos \dfrac {(ฮธ+ฮดฮธ + ฮธ)}{2} \) or \(ฮดy\) = 2\( cos (ฮธ+ \dfrac{ฮดฮธ}{2}) \) . \( sin( \dfrac {ฮดฮธ}{2}) \)โˆด \(\dfrac {ฮดy}{ฮดฮธ}\) = 2 \( \dfrac { cos (ฮธ+ \dfrac{ฮดฮธ}{2}) sin( \dfrac {ฮดฮธ}{2}) } {ฮดฮธ}\)or \(\dfrac {ฮดy}{ฮดฮธ}\) = 2 \(cos (ฮธ+ \dfrac{ฮดฮธ}{2}) \) \( \dfrac { sin( \dfrac {ฮดฮธ}{2}) } {ฮดฮธ}\) โˆด \(\dfrac […] READ MORE...

Derivative of cosฮธ using the First Principle

Derivative of \( cosฮธ \) using the First Principle Let \(y\) = \( cosฮธ \) โˆด \(y + ฮดy\) = \( cos(ฮธ + ฮดฮธ) \) โˆด \(ฮดy\) = \( cos(ฮธ + ฮดฮธ) \) - \( cosฮธ \)From Trigonometry , we have \( cos(A-B) \) = -2.\( sin \dfrac {(A+B)}{2} \).\( sin \dfrac {(A-B)}{2} \)Using the above rule, we get\(ฮดy\) = -2.\( sin \dfrac {(ฮธ+ฮดฮธ + ฮธ)}{2} \).\( sin \dfrac {(ฮธ+ฮดฮธ - ฮธ)}{2} \) or \(ฮดy\) = -2\( sin (ฮธ+ \dfrac{ฮดฮธ}{2}) \) . \( sin( \dfrac {ฮดฮธ}{2}) \)โˆด \(\dfrac {ฮดy}{ฮดฮธ}\) = -2 \( \dfrac { sin (ฮธ+ \dfrac{ฮดฮธ}{2}) sin( \dfrac {ฮดฮธ}{2}) } {ฮดฮธ}\)or \(\dfrac {ฮดy}{ฮดฮธ}\) = -2 \(sin (ฮธ+ \dfrac{ฮดฮธ}{2}) \) \( \dfrac { sin( \dfrac {ฮดฮธ}{2}) } {ฮดฮธ}\) โˆด \(\dfrac […] READ MORE...