Help with limits and derivatives?
lim x->2 (x^3-3x^2+2x)/(x^2-x-6) substitute x=2, we have 0 / -4 = 0 --------------------------------------... use the definition of first principle to find the gradient of the given curve at the point indicated f(x)= -x^2+3x where x=3 f(x) = -x^2 + 3x f(x+h) =-(x+h)^2 + 3(x+h) f(x+h)-f(x) = [-(x^2-2xh-h^2) + 3x+3h ] - [-x^2+3x] f(x+h)-f(x) = -x^2-2xh-h^2+3x+3h +x^2 -3x f(x+h)-f(x) = -2xh-h^2+3h [f(x+h)-f(x)] /h = -2x+h+3 lim h-->0 [f(x+h)-f(x)] /h = -2x+3 f'(x) = -2x+3 at x=3 , f'(3) = (-2)(3) + 3 = -3 f(3) = -3^2+3(3) =0 Equation of tangent line at (3,0) y-0 = -3(x-3) y=-3x+9 --------------------------------------... y=2x+7 f(x)=2x+7 f(x+h)=2(x+h)+7 f(x+h)-f(x)= 2x+2h+7 - 2x-7 f(x+h)-f(x) = 2h [f(x+h)-f(x)]/h = 2 lim h-->0 [f(x+h)-f(x)]/h = 2 y' = 2 --------------------------------------... f(x) = √(3x-2) f(x) = sqrt(3x-2) f(x+h) = sqrt(3x+3h-2) f(x+h)-f(x) = [sqrt(3x+3h-2)-sqrt(3x-2)] Multiply and divide by [sqrt(3x+3h-2)+sqrt(3x-2)] f(x+h)-f(x) = [sqrt(3x+3h-2)-sqrt(3x-2)] [sqrt(3x+3h-2)+sqrt(3x-2)] / [sqrt(3x+3h-2)+sqrt(3x-2)] The numerator is of the form (a-b)(a+b)=a^2-b^2 a= sqrt(3x+3h-2) b= sqrt(3x-2) a^2-b^2 = (3x+3h-2)-(3x-2) = 3h lim h-->0 [f(x+h)-f(x)] / h = 3h /[ h [sqrt(3x+3h-2)+sqrt(3x-2)] ] = 3 / sqrt(3x-2)+sqrt(3x-2) = 3/ (2sqrt(3x-2) ) --------------------------------------... f(x) = 1/(2x+3)^2 f(x+h) = 1/(2x+2h+3)^2 f(x+h)-f(x) = 1/(2x+2h+3)^2 - 1/(2x+3)^2 f(x+h)-f(x) = [(2x+3)^2 - (2x+2h+3)^2] /(2x+2h+3)^2 (2x+3)^2 expand the numerator f(x+h)-f(x) = [ (4x^2+12x+9)-(4x^2+12x+9+8xh+12h+4h^2) ] / (2x+2h+3)^2 (2x+3)^2 f(x+h)-f(x) = [ -8xh-12h-4h^2) ] / (2x+2h+3)^2 (2x+3)^2 [f(x+h)-f(x)] / h = (-8x-12+4h) / [ (2x+2h+3)^2 (2x+3)^2] lim h-->0 [f(x+h)-f(x)] / h = (-8x-12+4h) / [ (2x+2h+3)^2 (2x+3)^2] y' = (-8x-12) / (2x+3)^4 --------------------------------------...
The more useful formula is V = 1/3 A*h where A is the area, because that's how it is derived. And its true for anything coming together at a point, irrespective of the shape at the bottom. It is the same formula for a square pyramid (square at the bottom) as a cone (circle on the bottom) or any shape at all. Its the three dimensional equivalent of the area of a triangle, which is A = 1/2 B*h, where B is the base. This fact continues to higher dimensions; the 4 dimensional hypervolume of a hypercone is 1/4 V * h, where V is the volume of the base. You can prove them all the same way using calculus. Volume of Cone Derivation Proof proves it for 3 dimensions; n dimensions is the same. Now, as to proving this without calculus. Archimedes did this using a "calculus like argument". He first used an argument concerning breaking shapes into shells to show it didn't matter what the shape of the base was, or where the shape converged to a point. He then found combinations of shapes which come to a point and form a cube, allowing him to calculate the constant 1/3 for the formula, as at Page on iit.edu
FOIL for (x+h)^3 What are the steps to expand this?
For the best answers, search on this site https://shorturl.im/oCDFd (x-2)^3 = (x-2)(x-2)(x-2) = (x^2 - 2x - 2x + 4) (x-2) = (x^2 - 4x + 4) (x-2) = (x^3 - 4x^2 + 4x) + (-2x^2 + 8x - 8) = x^3 - 6x^2 +12x -8
The difference quotient is thisf(x + h) - f(x)-------------------- hWhose limit when h tends to zero is what they want you to calculate. It is important because this is the theoretical definition of d f(x)-------- dxHere f(x) happens to be 2/x and then f(x+h) = 2/(x+h). Substitute x by (x+h)And the diff. Quotient is 2 2 2 x - 2 ( x+h) -------- - -------- --------------------x + h x x ( x+h)------------------------ = ----------------------- h h ------ 1Work out the upper and is 2x-2x-2h = = -2hThe central is h x (x+h) so - 2 h -2--------------- = --------------h x (x+h) x (x+h)This is the diff quot.taking now limit h->0 d 2 -2--- ( ------ ) = -------dx x x^2And that's all Take careSimon
Following are the ways to derive formulae for [math]cos(2x)[/math]Since [math]cos(A+B)=cosAcosB-sinAsinB[/math][math]cos(2x)=cos(x+x)=cosx×cosx-sinx×sinx[/math]Therefore, [math]cos(2x)=cos^{2}x-sin^{2}x[/math]This is the first formula.From this, we can further solve to get more formulae as follows.Since, [math]sin^{2}x+cos^{2}x=1[/math][math]cos(2x)=cos^{2}x-sin^{2}x=cos^{2}x-(1-cos^{2}x)=2cos^{2}x-1[/math]Therefore, [math]cos(2x)=2cos^{2}x-1[/math]This is the second formula.Further solving this, keeping in mind the identity [math]sin^{2}x+cos^{2}x=1[/math][math]cos(2x)=2cos^{2}x-1=2(1-sin^{2}x)-1=1-2sin^{2}x[/math]Therefore, [math]cos(2x)=1-2sin^{2}x[/math]This is the third formula.For the fourth formula, we will need the first formula i.e. [math]cos(2x)=cos^{2}x-sin^{2}x[/math]Which is nothing but [math]cos(2x)=\frac{cos^{2}x-sin^{2}x}{1}=\frac{cos^{2}x-sin^{2}x}{cos^{2}x+sin^{2}x}[/math]..(Since, [math]cos^{2}x+sin^{2}x=1[/math])Dividing numerator and denominator by [math]cos^{2}x[/math], we get,[math]cos(2x)=\frac{1-tan^{2}x}{1+tan^2x}[/math]This is the fourth formula.Listing down all the formulae again.[math]cos(2x)=cos^{2}x-sin^{2}x[/math][math]cos(2x)=2cos^{2}x-1[/math][math]cos(2x)=1-2sin^{2}x[/math][math]cos(2x)=\frac{1-tan^{2}x}{1+tan^2x}[/math]Hope this helps! You can also get such questions solved and get the detailed solution within seconds using the Scholr app by just uploading a picture of the question and also get to be a part of an ever-growing community of students.
Find the slope of the demand curve D(p) = 20 ÷ √(p - 1) at point (5, 10)?
The difference quotient at (5,10) is [20/√(5 + h - 1) - 20/√(5 - 1)] / h or after a bit of simplification 20((1/√(4 + h)) - 1/2)/h Combine into a single fraction: 20((1/√(4 + h)) - 1/2)/h = 20(2 - √(4 + h))/(2h√(4 + h)) Multiply numerator and denominator by (2 + √(4 + h)) and expand the numerator. You should get 20(-h)/[2h√(4 + h) (2 + √(4 + h))] Assuming h ≠ 0, this reduces to -10/[√(4 + h) (2 + √(4 + h))] This is perfectly well-defined for h = 0, and is equal to the desired slope. Does this bring back old memories?
For the function f(x)= x^2 +2x-4, find lim h to 0 f(h+4)- f(4)/(h)?
Edit: Oops, sorry I solved for derivative by the definition: [ f (x + h) – f(x) ] / h = [ (x + h)^2 + 2(x + h) - 4 - (x^2 + 2x - 4 ) ] / h Expand the expressions in the numerator and group like terms. = [ x^2 + 2 x h + h^2 + 2x + 2h - 4 - x 2 - 2x + 4] / h = [ 2 x h + 2h + h^2] / h = 2 x + 2 + h lim h->0 2x + 2 + h as h -->0 = 2x + 2 Hence: f '(x) = 2x + 2 Here is the solution you need: lim h->0 f(h+4)- f(4)/(h) f(h + 4) = (h + 4)^2 + 2(h + 4) - 4 = h^2 + 8h + 16 + 2h + 8 -4 = h^2 + 10h + 20 f(4) = 16 + 8 - 4 = 20 lim h->0 f(h+4)- f(4)/(h) = lim h->0 h^2 + 10h + 20 - 20/(h)= lim h->0 h^2 + 10h + 20 - 20/(h)= lim h->0 h^2 + 10h /(h)= lim h->0 h + 10 = 10
F(x) = -x² + 2x + 1? find and simplify f(x+h) -f(x)/h?
well i know [f(x+h) -f(x)]/h as being the first derivative of a function. note that as the derivative approaches its limit, h approaches 0. n h is a common denominator for all terms. so: 1. expand the function as u so correctly did: -(x + h)² + 2(x+h) + 1/h = [-x² - 2xh - h² + 2x + 2h + 1] / h = 2. Factor out h from each term in both the numerator n denominator n this is done because you want to cancel the h in the denominator with the h factored from the term in the numerator so that the denominator is equal to a number (1). h[-x²/h - 2xh/h - h²/h + 2x/h + 2h/h + 1/h] / h[1] = -x²/h - 2x -h + 2x/h + 2 + 1/h = 3. Since the derivative function approaches its limit as h approaches 0,replace h in the expression with 0 and that yields: -x²/0 - 2x -0 + 2x/0 + 2 + 1/0 = -2x + 2 and this expression is your simplified answer Note that any term divided by 0 is equal to infinity but for arithmetic convenience infinity is represented as 0 when doing calculations