what is the 10th derivative of x=6u^4-78u^9 and y=5u^2/45u^6 with respect to y?????
547629
assuming u is an unknown variable y=0
1055222
Is that e^y?\n\nCompute the differential of both sides\n\n7dxe^y + 7xe^ydy + 9x^2dx = 20y^3dy\n(20y^3 - 7xe^y)dy = (7e^y + 9x^2)dx\ndy/dx = (7e^y + 9x^2)/(20y^3 - 7xe^y)
945648
Take the derivative of f :\nf'(x) = 5x^4 (x+7)^4 + 4 (x+7)^3 x^5\nf'(x)= x^4 (x+7)^3 [ 5 (x+7) + 4x]\nf'(x)= x^4 (x+7)^3 [ 9x + 35]\n\nSo the derivative is equal to 0 for the values :\nx= -7\nx= - 35/ 9\nx = 0\n\nthe function is increasing in [-11;-7],\nthe function is increasing in [-35/9;+12],
921060
In general if you have x^n, the derivative is n * x^(n-1).\n\nFor x² the answer is 2x^1 or just 2x.\n\ny = x²\ndy/dx = 2x
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
676902
Incomplete question. Are you looking for the anti-derivative of x, a, or b???\n\nput dx, da, or db at the end of the equation so we'll know.
84751
I presume you mean the third derivative of V(t,D) with respect to t, not D. The first derivative of V(t,D) with respect to t is acceleration a,\n \na=dV(t,D)/dt=-D/(t^2)\n\nThe second derivative of V(t,D) with respect to t is the boost B,\n\nB=da/dt=2D/(t^3)\n\nThe third derivative of V(t,D) with respect to t is\n\ndB/dt=-6D/(t^4)
945033
Derivative is y=cos x -5. Substituting in 0 gives you the slope -4. Now plug in the coordinates (0,3) to get the line, which is y=-4x+3
775275
2y^3 + 2x^3 - 9xy = x^5\n\nBy implicit differentiation\n\n6y² dy/dx + 6x² - 9y - 9x dy/dx = 5x^4\nSo 3y(2y - 3) dy/dx = 5x^4 - 6x² + 9y\n\nSo dy/dx = (5x^4 - 6x² + 9y)/[3y(2y - 3)]
1138866
I'll lead you in the most important part, but it is up to you to complete it.\n\nDerivative of x . y = x'.y + x.y'\n\nThen, you will have:\n\nf'(x) = (2x . ln x^2) + x^2 (ln x^2)'\n\nContinue from here, the only problem that you may have is finding the derivative of ln y , which you can see in the link below.
706213
Wait a minute, did you want to know when the function is zero or when its derivative is zero?  \n\nFor derivative:\ny' = 2x + sin x, right?\n\nIt looks like there's one place where it's 0.  Requires and numerical solution.\n\nWell, it's 0 at x=0, right?\nI assume that x is in radians.  1 radian is about 57 deg.  As x decreases from 0 both parts become more negative.\n\nIt looks like x = 0 is the only zero of the dirivative.
764395
Hey, firstly I need this corrected: f(x)= Ix^2=5x+6I, because there should not be an equal sign after the power of 2.... I can't do this w/o knowing... if it is subtraction than I will tell you that from what I rembember about this stuff is that absolute value functions are continuous but not differentiable I think.... Because at the point, the f(c) equals the f(x) which means it is cont. but the derivative of f(x) approaching the vertex from the left side does not equal the derivative of the f(x) from the right side...whcih means it is NOT differentiable.... sorry I can't remember more buddy
1105592
y+dy=5(x+dx)^2 +1\n  y+dy=5(x² +2xdx +dx²) +1\n    y+dy=5x²+1 +5(2xdx+dx²)\n       dy=10 xdx as dx² is very small\n          dy/dx= 10x
347106
Okay, if you're actually serious, you REALLY should be paying more attention in class, but I'm gonna answer it for two points anyway.\n\nThe derivative of x^2 (with respect to x) is 2x.\n\nHave fun.
1049778
y=f(x)=((x-5)/(2x+1))^3\nIf y=f(x)^n then    :    y'=n*f(x)^(n-1)*f'(x)\nNow here:\ny' = 3 * ((x-5)/(2x+1))^2 * [(1*(2x+1)-2*(x-5)) / (2x+1)^2]\nSimplifying:\n\ny'=33*(x-5)^2/(2x+1)^4
773933
1-In general the derivative of ln|U| is U'/U. In your case U = 2x^x + 4x + 1. Using the fact that x^x = exp(xlnx) you can easily check that the derivative of \n\nU' = 2(1 + lnx)x^x + 4.\n\nThus the derivative of your function is\n\n[2(1+lnx)x^x+4]/[2x^x+4x+1].\n\n2- The derivative of U^a, where a is a real number, is aU^{a-1}U'. In your case U = x e^x and a=1/2. Since U' =(x+1)e^x, one has\n\n(sqrt(xe^x))' =1/2{(x+1)/sqrt(xe^x)}e^x = 1/2{(x+1)/sqrt(x)}exp(x/2).
49770
There is no closed form for the nth derivative of this function.  I symbolically calculated and factored the first 50 derivatives on MathCAD, and there was not a pattern.  Sorry, but you're out of luck....