Guide :

# find the derivative of x + 2y = 20

find the derivative of x2 + 2y = 20

## Research, Knowledge and Information :

### What is the derivative of x^2y^2? | Socratic

... ("some function")^2]=2x*("some function")^2+x^2*2("some function")*"the derivative of the ... So d/(dx)(x^2y^2)=2xy^2+x^2 ... What is the derivative of #x^2y ...

### Online Derivative Calculator • Shows All Steps!

... for example we write "5x" instead of "5*x". The Derivative Calculator has to detect these cases and insert the ... For each calculated derivative, ...

### How do you find the derivative of sqrt(x^2+y^2)? | Socratic

... derivation with respect to x at constant y is an example of partial derivative: del/(delx)sqrt(x^2 ... do you find the derivative of #sqrt(x ... find #(d^2y)/ (dx ...

### What is the derivative of 2Y - Answers.com

Answers.com ® WikiAnswers ® Categories Science Math and Arithmetic Calculus What is the derivative of 2Y? ... 5x = 2y + 20 x = 2/5y + 4 Solve for y: ...

### Find Derivative of y = x^x. Classic - analyzemath.com

Find Derivative of y = x x. A calculus tutorial on how to find the first derivative of y = x x for x > 0.

### Implicit Double Derivatives: x^2 + y ^ 2 = 25 - Free Math Help

2) Find the second derivative of y = x^2 y^3 + xy I actually have no clue how to find the second derviative. ... Implicit Double Derivatives: x^2 + y ^ 2 = 25 1) ...

### Find The Derivative Of The Function. Y=x^4+10/x^2 ... - Chegg.com

Answer to Find the derivative of the function. y=x^4+10/x^2 A. d^2y/dx^2=2+60/x^4 B d^2y/dx^2=2x-20/x^3 C d^2y ... Find the derivative of the function. y=x^4+10/x^2 ...

### Derivative Calculator: Solve Derivatives with Wolfram|Alpha

Online Derivative Calculator Solve derivatives with Wolfram|Alpha: ... ^2 - 3x^2)/(h) = 6 x`. The derivative is a powerful tool with many applications. For example, ...

### SOLUTIONS TO IMPLICIT DIFFERENTIATION PROBLEMS

SOLUTIONS TO IMPLICIT ... and the first derivative as a function of x and y is (Equation 1) . To find y ... and the second derivative as a function of x and y ...

### real analysis - Did I take the derivative correctly? \$x^y=y^x ...

Did I take the derivative correctly? \$x^y=y^x\$ ... 6,296 5 20 59. asked Nov 1 '13 at 20:49. ... Thus to properly compute the derivative, you have to rewrite \$y^x = e^ ...

## Suggested Questions And Answer :

### can you please explain how to to find dy/dx for the function x^2 y+ Y^2 x = -2

I need to find dy/dx of this function and evaluate the derivative at the point (2,-1) x^2 y+y^2 x = -2 solve the equation to be  = 0 x^2y + y^2x +2 = 0 find partial derivatives for dy and dx, the first term has two parts x^2y has two partial derivatives 2xy dx + x^2 dy the second term has 2 parts y^2x has 2 partial derivatives y^2 dx + 2yx dy and 2 has no derivative 2xy dx + y^2 dx + x^2 dy + 2yx dy = 0 2xy + y^2 dx = (-x^2 - 2xy) dy dy/dx = (2xy + y^2)/(-x^2 - 2xy) use the point (2,-1) dy/dx = [2*2*(-1) + (-1)^2]/[-1(2)^2 -2*2*(-1)] dy/dx = (-4 + 1) / (-4 + 4)  there is a 0 in the de=nominator therefore it is undefined

### the integral of xtanx dx

Consider the function tanx=a0+a1x+a2x^2+...+a(n)x^n where a(n) is the coefficient of x^n. We need to find a(n). We can do this by applying calculus (effectively Taylor's theorem). If we integrate tanxdx we get -ln(cosx). If we integrate the power series we get C+a0x+a1x^2/2+a2x^3/3+...+a(n)x^(n+1)/(n+1), where C is a constant of integration. This is a power series for -ln(cosx) or strictly, -ln|cosx|, because we can only take logs of positive numbers. Also cosx can only assume values between 0 and 1, so the log is negative, and we can write -ln|cosx| as ln|secx|. |secx| is always 1 or more. Going back to the expansion for tanx, we know tan0=0, so a0=0. Therefore ln|secx|=C+a1x^2/2+...+a(n)x^(n+1)/(n+1). We'll see this pop up later when we integrate xtanx by parts. The derivative of tanx is sec^2x=S(na(n)x^(n-1)) where S is the sum of n terms (from n=1), since a0=0. When x=0, sec^2x=1 and when x=(pi)/2, sec^2x=0. The only term for S not containing x is a1, so a1=1. So far the series for tanx is: x+S(na(n)x^(n-1)) for n>2. Not much to go on yet. The next derivative is 2sec^2tanx=S(n(n-1)x^(n-2)) for n>2 (differentiation by substitution: let u=secx; du=secxtanxdx; d/dx=d/du*du/dx=2u*secxtanx=2sec^2xtanx). When x=0, tan0=0 so this derivative is zero, making 2a2=0, so a2=0. The next derivative is 4sec^2xtan^2x+2sec^4x (differentiation by parts: u=2sec^2x, v=tanx; du=4sec^2xtanxdx, dv=sec^2xdx; d(uv)=vdu+udv=(tanx)(4sec^2xtanxdx)+(2sec^2x)(sec^2xdx)). This derivative is 2 when x=0, so 6a3=2 and a3=1/3 (from n(n-1)(n-2)a(n)x^(n-3) where n=3). The 4th derivative is 8tanxsec^3x+8sec^2tan^3+8sec^4xtanx, which is zero when x=0 and a4=0. The 5th derivative is: 8tanx(3sec^3xtanx)+8sec^3x(sec^2x)+8sec^2x(3tan^2sec^x)+8tan^3(2sec^2tanx)+8sec^4x(sec^2x)+8tanx(4sec^4tanx) This derivative is 16 when x=0. The relevant term is 120a5, so a5=16/120=2/15. tanx=x+x^3/3+2x^5/15+... xtanx=x^2+x^4/3+2x^6/15+... integrate: x^3/3+x^5/15+2x^7/105+... Another way of approaching the series is to use the power series for sinx and cosx because tanx=sinx/cosx. Just as we found the coefficients of the power series for tanx, we can do the same for sinx, when we get sinx=x-x^3/3!+x^5/5!-x^7/7!+... And cosx is derivative of sinc, so cosx=1-x^2/2!+x^4/4!-... Also tanx*cosx=sinx, so we can use this identity to derive the coefficients for tanx. (a0+a1x+a2x^2+...)(1-x^2/2!+x^4/4!-...)=x-x^3/3!+x^5/5!-...=a0+a1x+...+a1x-a1x^3/2!+a1x^5/4!-...+a2x^2-a2x^4/2!... By equating the coefficients for a particular power of x we can work out the unknown coefficients a(n). For example, because there are no even powers of x in the expansion of sinx, a0=0 (which we already discovered), a2, a4, etc. are all zero. to find a1, we need all terms involving x. Since a1x is the only one, a1=1; to find a3, we have -a1x^3/2=-x^3/6 so a1=1/3; a1x^5/24-a3x^5/2+a5x^5=x^5/120, 1/24-1/6+a5=1/120, a5=1/120-1/24+1/6=(1-5+20)/120=16/120=2/15, as we discovered earlier. What is a7? To get the coefficient of x^7 we need to combine x with x^6, x^3 with x^4, x^5 with x^2 and x^7. The coefficients are a1, a3, a5 and a7 from tanx; -1/6!, 1/4!, -1/2! from cosx; -1/5040 from sinx. -a1/720+a3/24-a5/2+a7=-1/5040; -1/720+1/72-1/15+a7=-1/5040; a7=1/720-1/72+1/15-1/5040=(7-70+336-1)/5040=272/5040=17/315. Now we return to integral xtanxdx. Let u=x, then du=dx; dv=tanxdx, then v=ln|secx|, as we discovered earlier. d(uv)=vdu/dx+udv/dx=ln|secx|dx+xtanxdx. So integral(xtanxdx)=xln|secx|-integral(ln|secx|dx)=xln|secx|-(x^3/3+x^5/15+2x^7/105+...)+C.

### first derivative test

After you find the first derivative, set it equal to zero that will give you the potential candiates for absolute extrema. Then construct a sign chart of the first derivative for those numbers where you pick a number less than or between those numbers and then plug them into for x in the first derivative. Solve the equation and see if their sign is positive or negative then. If it is negative then positive it is min. If it is positive then negative it is min. Now that is local extrema to find absolute extrema take the values for the local extrema and plug those into the ORIGINAL EQUATION and find out which one is the biggest and smallest values.

### find the nth derivative of y=log(1+x)

Question: find the nth derivative of y=log(1+x). Using yn as the nth derivative, y = ln(1+x) y1 := 1/(1+x) y2 := -1/(1+x)^2 y3 := 2/(1+x)^3 y4 := -6/(1+x)^4 y5 := 24/(1+x)^5 By observation, we can see that the nth derivative will be given by yn = (-1)^(n-1).(n-1)! / (1+x)^n, n >= 1

### Find the 1st and 2nd derivative of 3xy=4x+y^2

Find the 1st and 2nd derivative of 3xy=4x+y^2 Differentiate both sides of 3xy=4x+y^2 implicitly. 3y + 3xy' = 4 + 2yy'  ----- (1) y'(3x - 2y) = 4 - 3y y' = (4 - 3y)/(3x - 2y)  ---- 1st derivative To get the 2nd derivative, it would probably be simpler to implicitly differentiate eqn (1). 3y' + 3y' + 3xy'' = 2(y')^2 + 2yy'' y''(3x - 2y) = 2(y')^2 - 6y' = 2y'(y' - 3) substituting for y' into the above expression. y''(3x - 2y) = 2{(4 - 3y)/(3x - 2y)}*{(4 - 3y)/(3x - 2y) - 3} y''(3x - 2y) = 2{(4 - 3y)/(3x - 2y)}*{[(4 - 3y) - 3(3x - 2y)]/(3x - 2y)} y''(3x - 2y) = 2{(4 - 3y)}*{[4 - 3y - 9x + 6y]}/(3x - 2y)^2 y'' = 2{(4 - 3y)}*{4 + 3y - 9x}/(3x - 2y)^3  --- 2nd derivative

### limit trigonometry

(tanx-x)/(x-sinx)=(sinx-xcosx)/(xcosx-sinxcosx). As x approaches 0 the expression approaches 2. Why? To find out, let's use some calculus. Let's suppose that sinx=a[0]+a[1]x+a[2]x^2+a[3]x^3+...+a[n]x^n, a polynomial series in x, with real coefficients a[0], etc. We are going to use a value for x<1 so the series would converge. When x=0, sinx=0 so a[0]=0. The derivative of sinx is cosx, so cosx=a[1]+2a[2]x+3a[3]x^2+...+na[n]x^(n-1). When x=0 cosx=1, so a[1]=1 and sinx=x+a[2]x^2+... The second derivative of sinx is -sinx=2a[2]+6a[3]x+... When x=0, sinx=0, so 2a[2]=0 and a[2]=0 and sinx becomes x+a[3]x^3. Moving on to the next derivative which is -cosx we have 6a[3]+... But -cosx when x=0 is -1, so a[3]=-1/6, making sinx=x-x^3/6+... We could keep on going through more derivatives, but it's not necessary because we can establish right now what x-sinx tends to as x approaches zero; x-sinx=x^3/6. Now, let tanx=b[0]+b[1]x+b[2]x^2+... just like we did with sinx. We can see that b[0]=0 because tanx=0 when x=0. The first derivative of tanx=sec^2x=b[1]+2b[2]x+3b[3]x^2+... secx is the reciprocal of cosx which is 1 when x=0, so b[1], like a[1], is 1 and tanx=x+b[2]x^2. The derivative of sec^2x is the derivative of cos^-2x=-2cos-3x*(-sinx)=2tanxsec^2x=0, when x=0, making b[2]=0. We need to go to the next derivative to find b[3]. But 2tanxsec^2x=2tanx(1+tan^2x)=2tanx+2tan^3x, which differentiates to 2sec^2x+6tan^2xsec^2x, and this derivative is 2 when x=0. Therefore 2=6b[3] and b[3]=1/3 and tanx-x=x^3/3. So the limit as x approaches zero of (tanx-x)/(x-sinx)=(x^3/3)/(x^3/6)=2.

### Find the directional derivative of f(x; y; z) = 1/(√ x^2+y^2+z^2) at P : (3; 0; 4) in the direction of a = [1; 1; 1].

Find the directional derivative of f(x; y; z) = 1/(√ x^2+y^2+z^2) at P : (3; 0; 4) in the direction of a = [1; 1; 1]. The directional derivative is: Du.f(x,y,z) = fx(x,y,z).a + fy(x,y,z).b + fz(x,y,z).c,  where, f(x, y, z) = 1/(√ x^2+y^2+z^2) and u = (a, b, c) = (1, 1, 1) fx = -x/(x^2+y^2+z^2)^(3/2), fy = -y/(x^2+y^2+z^2)^(3/2), fz = -z/(x^2+y^2+z^2)^(3/2) Du.f(x,y,z) = fx(x,y,z).a + fy(x,y,z).b + fz(x,y,z).c Du.f(x,y,z) = fx(x,y,z).1 + fy(x,y,z).1 + fz(x,y,z).1 Du.f(x,y,z) = fx(x,y,z) + fy(x,y,z) + fz(x,y,z) Du.f(x,y,z) = -x/(x^2+y^2+z^2)^(3/2)  –  y/(x^2+y^2+z^2)^(3/2)  –  z/(x^2+y^2+z^2)^(3/2) Du.f(x,y,z) =( -x –  y  –  z) / (x^2+y^2+z^2)^(3/2) Du.f(x,y,z) = -(x +  y  +  z) / (x^2+y^2+z^2)^(3/2) At P(3, 0, 4) D = -(3 +  0  +  4) / (9 + 0 +16)^(3/2) D = -(7) / (125)

### to find nth derivative of tan ^ ( - 1 ) ( x / a ) with solution

Let y=tan^-1(x/a), then tan y = x/a. sec^2(y)dy/dx=1/a so dy/dx=1/(a(sec^2(y))=1/(a(1+tan^2(y))=1/(a(1+(x^2/a^2))=1/(a+x^2/a)=(a+x^2/a)^-1 Write y'=dy/dx for convenience, so y'' is the second derivative and y'=(a+x^2/a)^-1 or sec^2(y)y'=1/a. Let u=v=sec y and w=y', then sec^2(y)dydx=1/a can be expressed uvw=1/a. Therefore we can get the second derivative by differentiating further: d/dx(u.v.w.)=v.w.du/dx+u.w.dv/dx+u.v.dw/dx=0 (differentiation by parts) 2sec(y).y'.sec(y)tan(y)y'+sec^2(y).y''=0 2sec^2(y)tan(y)y'^2+sec^2(y)y''=0, which we can divide through by sec^2(y): tan(y)y'^2+y''=0 making y''=-tan(y)y'^2=-x/a((a+x^2/a)^-2) So far, there's no obvious pattern in the derivatives which would suggest the nth derivative. Let's continue by finding the 3rd derivative, using the derivatives of y rather than converting to expressions in x. y'''=d/dx(-tan(y)y'^2) To calculate d/dx(y'^2) we can differentiate by parts and the result is 2y'y''. If we differentiate the right side of this equation we get y'''=-sec^2(y)y'-2tan(y)y'y''=-(1+tan^2(y))y'-2tan(y)y'y''=-y'(1+tan^2(y)+2tan(y)y''). Or, since we know that sec^2(y)y'=1/a, we can substitute to simplify the third derivative: y'''=-1/a-2tan(y)y'y''. Also, y''=-tan(y)y'^2 and tan(y)=x/a, so a simpler expression for y'''=-1/a-(2x/a)y'y''. More to follow...

### find the nth term of y=(cosx)(cos2x)(cos3x)

Given cos (x) cos(2x) cos(3x) let d/dx cos x = -sin x d^2/dx^2 cos x = -cos x d^3/dx^3 cos x = sin x and then things repeat, you get the same sequence of derivatives after that. If n is odd the nth derivative is: (-1)^( (n+1)/2 ) sin x And if n is even (-1)^( n/2 ) cos x d/dx cos 2x = -2 sin 2x So the only difference is in this case we get an extra 2 each step... d^2/dx^2 cos 2x = -4 cos 2x d^3/dx^3 cos 2x = 8 sin 2x d^4/dx^4 cos 2x = 16 cos 2x The sequence is predictable, the sequence of -sin 2x, -cos 2x, sin 2x, cos 2x, coupled with the powers of two (2, 2*2, 2*2*2, 2*2*2*2, ... ). And if instead we started with cos 3x, then the sequence would have had coefficients of 3,3*3,3*3*3,etc So by this pattern, the nth derivative of cos ax would be: If n is odd, (-1)^( (n+1)/2 ) a^n sin x And if n is even (-1)^( n/2 ) a^n cos x " Understand more about Finding the Nth Term - http://www.mathcaptain.com/algebra/binomial-theorem.html

### find compelet derivative of y = ln(sinex))2 cos(tanx)

y = ln(sinex))2 cos(tanx) dy/dx=2(lnsinex*-sin(tanx)*secx^2+cos(tanx)*cosx*e/sinx)