Given f(x) =e^x, x ≥0 =x+1, x<0 Find f^'(0) .

Name: Problem 21, Chapter 4: Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry
Uploaded: 2022-06-07T10:01:04-08:00
Duration: 6 min 24 s
Channel: Souvik Ghosh
Description: Given f(x) =e^x, x ≥0 =x+1, x<0 Find f^'(0) .

Given f(x) =e^x, x ≥0 =x+1, x<0 ...

Key Concepts

Piecewise Functions

Piecewise functions are functions that are defined by different expressions depending on the part of the domain. They require careful handling at the boundaries where the definition changes, especially when determining properties like continuity and differentiability.

Definition of Derivative

The derivative of a function at a point is defined as the limit of the difference quotient as the increment approaches zero. This concept provides the instantaneous rate of change and relies on establishing the existence of a unique limit that describes the slope of the tangent line at that point.

One-Sided Derivatives

One-sided derivatives involve evaluating the limit of the difference quotient using values that approach the point of interest from one side only, either from the left or the right. This concept is particularly essential when analyzing functions defined piecewise, as it helps determine if the function has a consistent rate of change at the boundary between pieces.

Differentiability at a Point

Differentiability at a specific point means that a function has a well-defined derivative there, which generally requires that both the left-hand and right-hand derivatives exist and are equal. In the context of piecewise functions, checking differentiability involves ensuring that the function is smooth at the transition point, often coinciding with the function being continuous.

Transcript

00:01 The function here given is f s equals e to the bar x for all x greater than equals to 0 and is equal to x plus 1 for all x less than 0.

00:09 And we have to find f prime 0.

00:12 So for that, first we write the expression for the first derivative from the first principle, which is f prime x equals limit h tends to 0, if of x plus h minus f of x divided by h.

00:38 Now for h greater than 0, x plus h is greater than x and for h less than 0, x plus h is less than x.

00:53 So, when is x is equal to 0, we will have x plus h greater than 0.

01:05 That is when h is greater than 0 and x plus h is less than 0 when h is less than 0 so in the first case that is for x plus h greater than 0 we will have f of x plus h equals a to the power x plus h and for h less than 0 we will have f of x plus h equals x plus h plus 1 when h is less than 0 and we have that x tend into 0 and here also x tending to 0 so for the first case that is when h is greater than 0 we have f prime x equals limit h tends to 0 plus f of x plus h is e to the power x plus h minus e to the power x divided by h this equals to limit h tends to zero plus now use the power properties of exponentials and write it as e to the power x into e to the power h minus e to the power x divided by h this gives us limit h tends to 0 plus factored out e to the power x then we left with e to be per h minus 1 divided by h which equals limit h tends to 0 plus now e to the power x is independent of h so we can bring it outside of the scope of limit so we are left with a to the power h minus 1 by h now this term that is limit h tends to 0 plus e to the bar h minus 1 by h will be equal to 1 so you'll have e to the power x into 1 equals e to the power x so for h greater than 0 and when x is tends to 0 we will have f prime 0 equals e to the power 0 equals e to the now for h less than 0, we will have f prime x equals limit h tends to 0 minus.

04:13 For this, the expression of fo x plus h is x plus h plus 1 minus x plus 1 divided by h...

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