0
Add your answer:
Earn +20 pts
What is the derivative of xy-2y equals 1?
by implicit differentiation you have y+x*dy/dx-2dy/dx=0 solving for dy/dx you'll have dy/dx=y/(2-x) and solving for y in the original equation and plugging it back in, you'll get dy/dx=1/(-x^2 +4x-4) which is your final answer
What is differential equation of x-y equals xy?
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
How do you find the derivative of 3 times the square root of x?
Suppose the relation is equal to y then y=3x^1/2 dy/dx = 3*1/2*x^-1/2*1 dy/dx = 3/2*1/x^1/2*1 dy/dx = 3/2x^1/2 First we have retained the constant term i.e. 3 and then we have applied the power rule on x i.e. (nxn-1).
What is the derivative of cos x raised to the x?
cos(xx)?d/dx(cosu)=-sin(u)*d/dx(u)d/dx(cos(xx))=-sin(xx)*d/dx(xx)-The derivative of xx is:y=xx ;You have to use implicit derivation because there is no formula for taking the derivative of uu.lny=lnxxlny=xlnx-The derivative of lnx is:d/dx(lnu)=(1/u)*d/dx(u)-d/dx(uv)= u*dv/dx+v*du/dxTherefore:(1/y)*dy/dx=x*[(1/x)*d/dx(x)]+lnx(d/dx(x))-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1(1/y)*dy/dx=x*[(1/x)*(1)]+lnx(1)(1/y)*dy/dx=x*[(1/x)]+lnx(1/y)*dy/dx=(x/x)+lnx(1/y)*dy/dx=1+lnxdy/dx=y(1+lnx) ;Multiply y to both sidesdy/dx=xx(1+lnx) ;y=xx, so replace the y with xxd/dx(cos(xx))=-sin(xx)*[xx*(1+lnx)]d/dx(cos(xx))=-(1+lnx)*xx*sin(xx)(cosx)x?Again with the implicit derivation:y=(cosx)xlny=x*ln(cosx)(1/y)*dy/dx=x[d/dx(lncosx)]+lncosx(d/dx(x))(1/y)*dy/dx=x[(1/cosx)*(-sinx)(1)]+lncosx(1) ;The derivative of lncosx is (1/cosx)*d/dx(cosx). The derivative of cosx is (-sinx)*d/dx (x). The derivative of x is 1.(1/y)*dy/dx=x[(1/cosx)*(-sinx)]+lncosx(1/y)*dy/dx=x[-tanx]+lncosx(1/y)*dy/dx=-xtanx+lncosxdy/dx=y(-xtanx+lncosx) ;Multiply both sides by ydy/dx=(cosx)x(-xtanx+lncosx) ;y=(cosx)x, replace all y's with (cosx)xdy/dx=(cosx)x(-xtanx+lncosx)=(cosx)x-1(cosx*lncosx-xsinx)
What is the differentiation of y with respect to x for x equals y exponent y?
x = yy differentiate both sides with respect to x dx = (y * yy-1) dy dy/dx = y * yy-1 dy/dx = yy = x hence differentiate of y wrt x is x only
What is the derivative of xy-2y equals 1?
by implicit differentiation you have y+x*dy/dx-2dy/dx=0 solving for dy/dx you'll have dy/dx=y/(2-x) and solving for y in the original equation and plugging it back in, you'll get dy/dx=1/(-x^2 +4x-4) which is your final answer
What is differential equation of x-y equals xy?
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
Integral of sin square root x?
For ∫ sin(√x) dx let y = √x = x1/2 → dy = 1/2 x-1/2 dx → 2x1/2 dy = dx → 2y dy = dx → ∫ sin(x1/2) dx = ∫(sin y) 2y dy Now: ∫ uv dx = u∫v dx - ∫(u'∫v dx) dx → ∫(sin y) 2y dy = ∫2y sin y dy = 2y ∫sin y dy - ∫(2 ∫sin y dy) dy = -2y cos y + 2 sin y + C = 2 sin y - 2y cos y + C → ∫ sin(√x) dx = 2 sin(√x) - 2(√x) cos(√x) + C
What is the integral of x3 ex4 dx?
∫x3ex4 dx = 1/4ex4 + c To solve, let y = x4, then: dy = 4x3 dx ⇒ 1/4dy = x3 dx ⇒ ∫x3ex4 dx =∫ex4 x3 dx = ∫ey 1/4 dy = 1/4ey + c but y = x4, thus: = 1/4ex4 + c
How do you find the derivative of 3 times the square root of x?
Suppose the relation is equal to y then y=3x^1/2 dy/dx = 3*1/2*x^-1/2*1 dy/dx = 3/2*1/x^1/2*1 dy/dx = 3/2x^1/2 First we have retained the constant term i.e. 3 and then we have applied the power rule on x i.e. (nxn-1).
What is the derivative of cos x raised to the x?
cos(xx)?d/dx(cosu)=-sin(u)*d/dx(u)d/dx(cos(xx))=-sin(xx)*d/dx(xx)-The derivative of xx is:y=xx ;You have to use implicit derivation because there is no formula for taking the derivative of uu.lny=lnxxlny=xlnx-The derivative of lnx is:d/dx(lnu)=(1/u)*d/dx(u)-d/dx(uv)= u*dv/dx+v*du/dxTherefore:(1/y)*dy/dx=x*[(1/x)*d/dx(x)]+lnx(d/dx(x))-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1(1/y)*dy/dx=x*[(1/x)*(1)]+lnx(1)(1/y)*dy/dx=x*[(1/x)]+lnx(1/y)*dy/dx=(x/x)+lnx(1/y)*dy/dx=1+lnxdy/dx=y(1+lnx) ;Multiply y to both sidesdy/dx=xx(1+lnx) ;y=xx, so replace the y with xxd/dx(cos(xx))=-sin(xx)*[xx*(1+lnx)]d/dx(cos(xx))=-(1+lnx)*xx*sin(xx)(cosx)x?Again with the implicit derivation:y=(cosx)xlny=x*ln(cosx)(1/y)*dy/dx=x[d/dx(lncosx)]+lncosx(d/dx(x))(1/y)*dy/dx=x[(1/cosx)*(-sinx)(1)]+lncosx(1) ;The derivative of lncosx is (1/cosx)*d/dx(cosx). The derivative of cosx is (-sinx)*d/dx (x). The derivative of x is 1.(1/y)*dy/dx=x[(1/cosx)*(-sinx)]+lncosx(1/y)*dy/dx=x[-tanx]+lncosx(1/y)*dy/dx=-xtanx+lncosxdy/dx=y(-xtanx+lncosx) ;Multiply both sides by ydy/dx=(cosx)x(-xtanx+lncosx) ;y=(cosx)x, replace all y's with (cosx)xdy/dx=(cosx)x(-xtanx+lncosx)=(cosx)x-1(cosx*lncosx-xsinx)
deriative of x?
Let y = x Then the derivative is dy/dx = 1 Method Think of xto the power of ;1; y = x^1 Casually To derive a statement make the power number the coefficient. Then subtract one from the power number. Hence y = x^1 dy/dx = (1)x^(1-1) NB The '1-1' = 0' Hence dy/dx = (1)x^(0) NB Any value with a power of 'zero' is equal to one. Hence dy/dx = (1)(1) = 1 dy/dx = 1 NNB THe algebraic proof of derivatives is some 'heavy' algebra; too long for a short answer here, Hope that helps!!!!
What is the derivative of x-y?
The partial derivative in relation to x: dz/dx=-y The partial derivative in relation to y: dz/dy= x If its a equation where a constant 'c' is set equal to the equation c = x - y, the derivative is 0 = 1 - dy/dx, so dy/dx = 1
What is the differentiation of y with respect to x for x equals y exponent y?
x = yy differentiate both sides with respect to x dx = (y * yy-1) dy dy/dx = y * yy-1 dy/dx = yy = x hence differentiate of y wrt x is x only
Differential co-efficient of log tan x is?
y = ln(tan(x)) u = tanx y =ln(u) dy/du = 1/u du/dx = sec2(x) dy/dx = dy/du * du/dx = sec2(x)/tan(x)
What is the derivative of -x-y?
If y is a function of x, that is y=f(x), then the derivative of x-y is 1-y' or 1-dy/dx (where y' or dy/dx is the differential coefficient of y with respect to x).
How do you differentiate exp exp x?
Use the "chain rule" of differentiation: y=exp(exp(x)) taking ln both side in y=e x (1/y)dy/dx=e x dy/dx=y*e x dy/dx=exp(x+exp(x))
