What is [math]\frac{d}{dx}[/math] and what is the difference between [math]\frac{d}{dx}[/math] and [math]\frac{dy}{dx}[/math]? What is the chain rule?
d/dx is somewhat of an operator which means “differentiate whatever is next to it with respect to x". You can put anything in front of it, from a variety of functions to constant numbers to functions in other variables etc.dy/dx specifically means that you are differentiating y with respect to x. Y again can be anything.The chain rule is important. If you have a composite function, you'll have to differentiate it in such a way that you have to handle both the change of output as well as the change of its input because the input will be another function which will be changing with respect to x.Chain rule sounds dangerous but is actually quite easy. First differentiate the outermost function, keeping the interiors intact. Move onto the next function and do the same thing while multiplying it with the previous result. Keep repeating this until you run out of functions to differentiate.
Find dy/dx using implicit differentiation give x^2y+y^4-3x=8?
This is my first year in Calc and it's killing me already, if you can kindly show the steps to solve this it would greatly aide me in understanding the concept better.
Find (dy/dx) for y= (x^3) (square root 2x +1)?
[(3x^2)(sqrt2x+1)-(x^3)(1/2sqrt2x+1)]/2x...
USe the chain rule to compute the derivative dy/dx for the given value of x?
First find dy/du for y= 3u^4-4u+5. This is like finding dy/dx except instead of x you have u. Use the basic power rule for this. dy/du = 12u^3 - 4 If u is x^3-2x-5 and you are trying to find the derivative at x=2, figure out what u equals when x is 2 by plugging in 2 for x in x^3-2x-5 . 2^3 - 2*2 - 5 = 8 - 4 - 5 = -1 u is -1 when x is 2. Now plug in -1 for u in dy/du 12* (-1)^3 - 4 = -12 - 4 = -16 dy/du = -16 Next find du/dx by finding the derivative of u=x^3-2x-5. This is the same as if the u were a y. du/dx = 3x^2 - 2 Since you are finding the derivative at x = 2, plug in 2 for x into du/dx to find du/dx at 2. 3*2^2 - 2 = 10 du/dx = 10 Here is where the chain rule comes into play: dy/dx = dy/du * du/dx Notice how the du seems cancel from the denominator and numerator when multiplied, leaving dy/dx. Although this cancellation should not be seen that way, it is okay to think like that for this problem. You found dy/du to be -16 and du/dx to 10 (both for x=2). Multiply this and you will get -160
How do I find the slope of a tangent to the curve y= (x^3-7) ^5 at x=2?
[math]y=(x^3-7)^5=f(x)[/math]The slope of a line tangent to [math]f(x)[/math] in a point [math]a[/math] is given by [math]f'(x)[/math].First, you have to find the derivative of [math]f(x)[/math].Because it's in the form [math]g[/math][math](h(x))[/math], we can use the Chain Rule with [math]g (x)=x^5[/math] and [math]h(x)=x^3-7[/math]The Chain Rule says that [math]\frac {d}{dx}g (h (x))=h'(x)g'(h (x))[/math][math]\implies f'(x)=(3x^2)(5 (x^3-7)^4)[/math][math]\implies f'(x)=15x^2 (x^3-7)^4[/math]Now you can just evaluate [math]f'(2)[/math][math]f'(2)=60[/math]The equation of a line tangent to a function [math]f(x)[/math] in a point [math]a[/math] is given by: [math]r:y-f (a)=f'(a)(x-a)[/math]You can just substitute with [math]a=2[/math] and you get:[math]r:y=60x-119[/math]Hope this helps :)
Need some calculus help please?
1. y = (t^2 + 1)/(1 - t^2) dy/dt = ((1 - t^2)(2t) - (-2t)(T^2 + 1))/(1 - T^2)^2 = 4t/(1 - t^2)^2 2. dy/du = 5u^4 - 6u + 6 du/dx = 2x dy/dx = (dy/du)(du/dx) = 5(x^2 - 1)(2x) - 6(x^2 - 1)(2x) + 6(2x) Plug in x = 1: Answer: 5(0)(2) - 6(0) + 6(2) = 12 3. y = x^2 - 3x + 5 dy = 2x dx - 3 dx = 2(5)(.3) - 3(.3) = 3 - 0.9 = 2.1 (Test : f(5) = 15 and f(5.3) = 17.29 ==> f(5.3) - f(5) = 2.29)
Find dy/dx by implicit differentiation. 3x^3 + x^2y - xy^3 = 3?
Use the Chain rule, and every time you run into a y --- use the chain rule + add a dy/dx into it.. 9x^2 + (x^2*dy/dx+y*2x) - x*3*y^2*dy/dx+y^3*(1) = 0 x^2 * dy - x^3y^2 * dy . ----- .------ = .-9x^2 - y^3 - 2xy . dx ................... dx factor out dy/dx and solve for it. dy ---- ( . x^2 - x^3y^2 .) = -9x^2 - y^3 - 2xy dx dy . -9x^2-y^3 - 2xy ---- = ----------------------- dx . x^2 - x^3y^2 dy/dx =(-9x^2-y^3-2xy) / (x^2-x^3*y^2) periods are to hold spaces
Find the derivative of the function. y = ((x^2 + 4) / (x^2 - 4))^3?
let u = x^2 + 4 and v = ( x^2 - 4 )^3 then y = u/v dy/dx = [ v du/dx - u dv/dx ] / v^2 du/dx = 2x dv/dx = 3 ( x^2 - 4 )^2 * 2x = 6x ( x^2 - 4 )^2 then dy/dx = [ (x^2 - 4 )^3 * 2x - ( x^2 + 4 ) * 6x * ( x^2 - 4 )^2 ] / ( x^2 - 4 )^6 dy/dx = 2x ( x^2 - 4 )^2 [ x^2 - 4 - 3x^2 - 12 ] / ( x^2 - 4 )^6 dy/dx = -4x ( x^2 + 8 ) / ( x^2 - 4 )^4
Calculus problems due tommorow, can somebody help me please?
1) f(x) = cos^2 (sin(2x)) Use the chain rule: [2] * [cos(2x)] * [-sin(sin(2x))] * [2cos(sin(2x))] = -4cos(2x) * sin(sin(2x)) * cos(sin(2x)) f'(pi/8) = -4cos(pi/4) * sin(sin(pi/4)) * cos(sin(pi/4)) =-4 [sqrt(2)/2 * sin(sqrt(2)/2) * cos(sqrt(2)/2)] =-1.396911997 2) dy/dx = (x^2) / (y^3) Separate the variables: dy * y^3 = dx * x^2 y^4 /4 = x^3 /3 + C If x = 3 and f(3) = 3: 3^4 /4 = 3^3 /3 + C 81/4 = 9 + C C = 45/4 y^4 /4 = x^3 /3 + (45/4) y^4 = (4/3)x^3 + 45 y = [ (4/3)x^3 + 45 ] ^ (1/4) 3) [2y^2 * x^3] - [5x^2 * y] = 18 Use implicit differentiation: [2y^2 * 3x^2 + 4y *(dy/dx) *x^3] - [5x^2 * (dy/dx) + 10x * y] = 0 [2y^2 * 3x^2 - 10x * y] + [4y *(dy/dx) *x^3 - 5x^2 * (dy/dx)] = 0 [6(xy)^2 - 10xy] + (dy/dx)* [4yx^3 - 5x^2 ] = 0 (dy/dx)*[4yx^3 - 5x^2 ] = -[6(xy)^2 - 10xy] (dy/dx) = [10xy - 6(xy)^2] / [4yx^3 - 5x^2 ] At (1,2), the dy/dx is the slope of the tangent line: (dy/dx) = [10*1*2 - 6(1*2)^2] / [4*2*(1)^3 - 5(1)^2 ] =[20 - 24] / [8 - 5] = -4/3 I get a thumbs-down for effort =)