Find the derivative as function using the the definition. f(x) = Sqrt(2x+3)?
Hello, f(x) = √(2x + 3) let's apply the definition f '(x) = lim {[f(x + h) - f(x)] /h}: ............. ......... .......... .......... ...h→0 f '(x) = lim { {√[2(x + h) + 3] - √(2x + 3)} /h} = ...........h→0 lim {[√(2x + 2h + 3) - √(2x + 3)] /h} = h→0 of course the limit takes the form 0/0, so let's mulripy numerator and denominator by √(2x + 2h + 3) + √(2x + 3), yielding a difference of squares in the numerator: lim { {[√(2x + 2h + 3) + √(2x + 3)] [√(2x + 2h + 3) - √(2x + 3)]} /{h [√(2x + 2h + h→0 3) + √(2x + 3)]} } = lim { {[√(2x + 2h + 3)]² - [√(2x + 3)]²} /{h [√(2x + 2h + 3) + √(2x + 3)]} } = h→0 lim {[(2x + 2h + 3) - (2x + 3)] /{h [√(2x + 2h + 3) + √(2x + 3)]} } = h→0 lim {(2x + 2h + 3 - 2x - 3) /{h [√(2x + 2h + 3) + √(2x + 3)]} } = h→0 lim {(2h) /{h [√(2x + 2h + 3) + √(2x + 3)]} } = h→0 (simplifying) lim {2 /[√(2x + 2h + 3) + √(2x + 3)]} = h→0 (letting h approach zero) lim {2 /{√[2x + 2(→0) + 3] + √(2x + 3)} } = h→0 lim {2 /{√[2x + (→0) + 3] + √(2x + 3)} } = h→0 2 /[√(2x + 3) + √(2x + 3)] = 2 /[2√(2x + 3)] = 1 /√(2x + 3) I hope it's helpful
Find the derivative of the function using the definition of derivative.?
a) By the definition of the derivative, we get: f'(x) = lim (h-->0) [f(x + h) - f(x)] / h = lim (h-->0) [5.5(x + h)^2 - (x + h) + 4.2 - (5.5x^2 - x + 4.2)] / h = lim (h-->0) [5.5(x^2 + 2xh + h^2) - x - h + 4.2 - 5.5x^2 + x - 4.2) / h = lim (h-->0) (5.5x^2 + 11xh + 5.5h^2 - x - h + 4.2 - 5.5x^2 + x - 4.2) / h = lim (h-->0) (5.5h^2 + 11xh - h) / h = lim (h-->0) [h(5.5h + 11x - 1)] / h = lim (h-->0) 5.5h + 11x - 1 = 5.5(0) + 11x - 1 = 11x - 1 So you are correct! b) If there are domain restrictions, then there are values that the function will be undefined at. However, this function is always defined. So the domain will be (-INFINITY, INFINITY). c) Same with this one. The function is always defined. So the domain is (-INFINITY, INFINITY) as well. I hope this helps!
Find the derivative of the function using the definition of derivative.?
I will support you with the by-product. Put to your calculator or graph it and you're going to see the area is all real numbers. F'(t)= x+delta x- x / delta x lim delta x --> 0 f(t)= 8t- 3t^2 f'(t) LDx-->0 = eight(t+delta t)- 3(t+delta t)^2 -[8t-3t^2] / delta t = 8t+8delta t-3(t^2+2t(delta t)+ (delta t)^2 -8t+3t^2 / delta t =eight(delta t) + 6t(delta t) +3(delta t^2) ----------------------------------------... = eight + 6t + 3delta t = 8+6t+3(0)= 8+6t delta t
Find the derivative of the function without using the Quotient Rule.?
lim h->0 (f(x + h) - f(x)) / h 9 * ((x + h)^(5/2) - x^(5/2)) / h => 9 * ((x + h)^(5/2) - x^(5/2)) * ((x + h)^(5/2) + x^(5/2)) / (h * ((x + h)^(5/2) + x^(5/2)) => 9 * ((x + h)^5 - x^5) / (h * ((x + h)^(5/2) + x^(5/2)) => 9 * (x^5 + 5x^4 * h + 10x^3 * h^2 + 10x^2 * h^3 + 5x * h^4 + h^5 - x^5) / (h * ((x + h)^(5/2) + x^(5/2))) => 9 * h * (5x^4 + 10x^3 * h + 10x^2 * h^2 + 5x * h^3 + h^4) / (h * ((x + h)^(5/2) + x^(5/2))) => 9 * (5x^4 + 10x^3 * h + 10x^2 * h^2 + 5x * h^3 + h^4) / ((x + h)^(5/2) + x^(5/2)) h goes to 0 9 * (5x^4 + 0 + 0 + 0 + 0) / ((x + 0)^(5/2) + x^(5/2)) => 45x^4 / (x^(5/2) + x^(5/2)) => 45 * x^4 / (2 * x^(5/2)) => (45/2) * x^(4 - 5/2) => (45/2) * x^(3/2)
How do i use the definition of derivative to find velocity?
I assume you mean f (t) = - 16t^2 + 48t + 64 and not - 16t^x + 48t + 64. By definition, velocity at t = 1 is f '(1) = lim (t → 1) [f (t) - f (1)] / (t - 1) = lim (t → 1) [- 16t^2 + 48t + 64 - (-16 + 48 + 64)] / (t - 1) = lim (t → 1) [- 16t^2 + 48t + 64 + 16 - 48] / (t - 1) = - lim (t → 1) [16(t^2 - 1) - 48(t - 1)] / (t - 1) = - lim (t → 1) [16 (t - 1) (t + 1) - 48 (t - 1)] / (t - 1) = - lim (t → 1) [16 (t + 1) - 48] = - (32 - 48) = 16.
Finding the derivative without an equation?
For a continuous function, locate points where f(x) = 0. These are the solutions to the graph you see. Let's say it has two solutions a and b. These would be the results of factoring the equation. So: x = a and a = b x - a = 0, a - b = 0 The equation of the graph is then (x-a)(x-b). Expand this using FOIL. f(x) = x² - ax - bx + ab f(x) = x² - (a+b)x + ab Take the derivative of f(x), and follow the usual directives to sketch a graph. Otherwise, the value of the derivative for a point on the graph is the slope of the line tangent to the graph. All points where slope = 0 is a solution to dF/dx = 0. Do the same as above by reverse engineering the factors and finding the equation
If you need only the derivative value on some specific point, you can view your function as if its variable is defined on the complex plane, which is to extend the definition of the function from [math]\mathbb{R}[/math] to [math]\mathbb{C}[/math].If the function [math]f(x)[/math] (extended to [math]f(z)[/math]) is "well-defined" on [math]\mathbb{C}[/math], i.e. it extends to a complex-analytic function (also called holomorphic function) Holomorphic function, then its [math]n[/math]-th derivative's value on [math]a[/math], which is [math]f^{n} (a)[/math], is given by Cauchy's integral formula. This formula unifies derivatives and integrals on the complex plane.For details please view this page: Cauchy's integral formula.Sorry that I might have use wrong terms on mathematics. I hope the idea is clear.
I think you meant to ask this as in the mathematical definition of a derivative as opposed by the financial derivative definition. (This question has been posted in the Derivatives (finance) topic). Anyway I think you can find the anser to this one in any site that deals with basic derivation definition, where the derivative of a mathematical function is the slope of the line which touches the function at any given point. The definition of the slope of this tangent line is obtained by inputting 2 values in the function and dividing by the difference in these values, then calculating it for the smallest possible difference in those values, for a particular point. The smallest possible difference will be the limit of this division when the difference between the 2 input values tends to zero. Say you input the values 2 and 4 in this function you will obtain the results 24 (4 x 2^2+8) and 72 (4 x 4^2+8). Dividing the difference between these values (72–24) by the difference between the inputted values (4–2) you get the rough slope of (72–24)/(4–2) = 48/2 = 24. I can´t write the symbols required for showing limit calculations using the keyboard, but the notion is that as you approach the values of the two input values to the limit of zero the result of this division will be the absolute slope of the line at any particular point. If you use 2 and 3 you will obtain (44–24)/1 = 20 as the slope. If you use very close values like 2 and 2.01 you get (24.1604–24)/(0.01) which gives 16.04. So as you begin to reach the ultimate closeness you also begin to approach the actual derivative which at this point 2 which is 16. The derivative function of f(x) = 4x^2+8 is f´(x)=8x, which, for x=2, yields the result 16 we just saw above. The formal limit equation is easily found by searching for ‘derivative function limit slope explanation’ on any search engine.
Let f x( ) = sin x . Prove that f '(x ) = cos x .f'( x) = lim (f (x+ h) - f (x))/h as h tends to 0 (zero)= lim (sin( x + h) - sin( x ))/h as h tends to 0 (zero)= lim (sin xcos h + cos xsin h - sin x)/ h as h tends to 0 (zero)( by Sum Identity for sine )= lim (( sin x cosh-b sin x )+ cos x sin h)/ h as h tends to 0 (zero)(Group terms with sin x)= lim ( sin x ( cosh -1 )+ cos x sin h )/hNow, group expressions containing h.= lim [sin x{ (cos h -1)/ h }tends to 0 as h tends to 0 (zero) + cos x {sinh/ h} tends to 1 as h tends to 0 (zero)]= cos x .Q.E.D.
This is a composite function, so use the chain rule. It also involves addition twice. Remember the derivative of a square root is 1/twice the root (this follows from using power 1/2. So the first step in the chain rule is simply [math]\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}[/math]. You then multiply by the derivative of the inside function which is 1 + a similar derivative with one less square root. This in turn uses the chain rule and the derivative of a sum. You can write it all down in one line because the later steps are just multiples of what you have so far. I don't know if the whole mess simplifies. But you might as well leave it as it is. I don't fancy trying to write this out in Latex in one expression, but the next factor is [math]1 + A\frac{1}{2\sqrt{x+\sqrt{x}}}[/math] where [math]A = 1 + \frac{1}{2\sqrt{x}}[/math] with, of course, parentheses around them.