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## Related Questions

###### Asked in Math and Arithmetic, Calculus

### What is the antiderivative of sine squared?

∫sin2x dx
Use the identity sin2x = ½ - ½(cos2x)
∫[½ - ½(cos2x)] dx = ∫½ dx - ∫½(cos2x) dx
Let's split it up into ∫½ dx and ∫½(cos2x) dx
∫½ dx = x/2 (we'll put the constant in at the end)
∫½(cos2x) dx (Use u substitution with u=2x and du = 2 dx)
∫cosu ¼du = ¼∫cosu du = ¼sinu + c = ¼sin2x (remember to
resubstitute)
Subtract the two parts and add a constant
x/2 - ¼(sin2x) + c
This is also equivalent to: ½(x - sinxcosx) + c

###### Asked in Economics

### How do you solve derivatives?

There are several ways to solve for the derivative:
You can use the limit definition: limhà0=(f(x+h)-f(x))/h.
You can use the limit at a point definition (when you actually
have a point, or if not use #1): limx->a=(f(x)-f(a))/(f-a).
Use the short cuts:
The derivative of a constant like 2 or pi is 0.
For things raised to a power: d/dx(xn)=nxn-1 where n is any
real number.
Multiplying by a constant, multiply the derivative of u by the
constant c: d/dx(cu)=c*du/dx
Adding, you add the derivative of the first part and the
derivative of the second: d/dx(u+v)=du/dx+dv/dx
Subtracting, you subtract the derivative of the first part and
the derivative of the second: d/dx(u-v)=du/dx-dv/dx
(Multiplying and dividing are more complicated, and you'll just
have to memorize their formulas)
Multiplying, the formula is: d/dx(uv)= u*dv/dx+v*du/dx
Dividing, the formula is: d/dx(u/v)=(v*du/dx-u*dv/dx)/(v2)
d/dx(v/u)=(u*dv/dx-v*du/dx)/(u2)

###### Asked in Math and Arithmetic, Algebra, Calculus

### What is Dydx of x(x-10)?

Assuming y = x(x - 10)
Either use the product rule:
u = x
v = (x - 10)
→ y = x(x - 10) = uv
d/dx (uv) = v du/dx + u dv/dx
→ dy/dx = d/dx x(x - 10)
= (x - 10)(d/dx x) + (x)(d/dx (x - 10))
= (x - 10)(1) + (x)(1)
= x - 10 + x
= 2x - 10
or expand the brackets and differentiate
dy/dx = d/dx x(x - 10)
= d/dx x² - 10x
= 2x - 10

###### Asked in Math and Arithmetic

### Derivative of 5ex plus 2?

5ex+2?
d/dx(u+v)=du/dx+dv/dx
d/dx(5ex+2)=d/dx(5ex)+d/dx(2)
-The derivative of 5ex is:
d/dx(cu)=c*du/dx where c is a constant.
d/dx(5ex)=5*d/dx(ex)
-The derivative of 2 is 0 because it is a constant.
d/dx(5ex+2)=(5*d/dx(ex))+(0)
d/dx(5ex+2)=5*d/dx(ex)
-The derivative of ex is:
d/dx(eu)=eu*d/dx(u)
d/dx(ex)=ex*d/dx(x)
d/dx(5ex+2)=5*(ex*d/dx(x))
-The derivative of x is:
d/dx(xn)=nxn-1
d/dx(x)=1*x1-1
d/dx(x)=1*x0
d/dx(x)=1*(1)
d/dx(x)=1
d/dx(5ex+2)=5*(ex*1)
d/dx(5ex+2)=5*(ex)
d/dx(5ex+2)=5ex
5ex+2?
d/dx(cu)=c*du/dx where c is a constant.
d/dx(5ex+2)=5*d/dx(ex+2)
-The derivative of ex+2 is:
d/dx(eu)=eu*d/dx(u)
d/dx(ex+2)=ex+2*d/dx(x+2)
d/dx(5ex+2)=5*(ex+2*d/dx(x+2))
-The derivative of x+2 is:
d/dx(u+v)=du/dx+dv/dx
d/dx(x+2)=d/dx(x)+d/dx(2)
d/dx(5ex+2)=5*[ex+2*(d/dx(x)+d/dx(2))]
-The derivative of x is:
d/dx(xn)=nxn-1
d/dx(x)=1*x1-1
d/dx(x)=1*x0
d/dx(x)=1*(1)
d/dx(x)=1
-The derivative of 2 is 0 because it is a constant.
d/dx(5ex+2)=5*[ex+2*(1+0)]
d/dx(5ex+2)=5*[ex+2*(1)]
d/dx(5ex+2)=5*[ex+2]
d/dx(5ex+2)=5ex+2

###### Asked in Video Games

### What are some cheats for Free Realms?

srry guys, i only know a few, but here they are...
FROGGY turns you into a frog
ICE CREAM SURPRISE surrounds you with snow flakes
SANDWICH makes you smaller...(i would recommend it, it is
halarious)
and
STRAWBERRIES gives you some sort of straw berry hat...
XD XD XD XD XD XD XD XD XD XD XD XD DX DX DX DX DX DX DX DX DX
DX DX DX

###### Asked in Calculus

### What is the derivative of x5lnx?

x5lnx?
d/dx (uv)=u*dv/dx+v*du/dx
d/dx (x5lnx)=x5*[d/dx(lnx)]+lnx*[d/dx(x5)]
-The derivative of lnx is:
d/dx(lnu)=(1/u)*[d/dx(u)]
d/dx(lnx)=(1/x)*[d/dx(x)]
d/dx(lnx)=(1/x)*[1]
d/dx(lnx)=(1/x)
-The derivative of x5 is:
d/dx (xn)=nxn-1
d/dx (x5)=5x5-1
d/dx (x5)=5x4
d/dx (x5lnx)=x5*[1/x]+lnx*[5x4]
d/dx (x5lnx)=[x5/x]+5x4lnx
d/dx (x5lnx)=x4+5x4lnx

###### Asked in Math and Arithmetic, Algebra, Calculus

### What is the Second derivative of 3.9625.lnx?

3.9625lnx?
The first derivative is:
d/dx(cu)=c*du/dx where c is a constant.
d/dx(3.9625lnx)=3.9625*d/dx(lnx)
-The derivative of lnx is:
d/dx(lnu)=(1/u)*d/dx(u)
d/dx(lnx)=(1/x)*d/dx(x)
d/dx(3.9625lnx)=3.9625*[(1/x)*d/dx(x)]
-The derivative of x is:
d/dx(xn)=nxn-1
d/dx(x)=1*x1-1
d/dx(x)=1*x0
d/dx(x)=1*(1)
d/dx(x)=1
d/dx(3.9625lnx)=3.9625*[(1/x)*1]
d/dx(3.9625lnx)=3.9625*(1/x)
d/dx(3.9625lnx)=3.9625/x
The second derivative of 3.9625lnx is the derivative of
3.9625/x=3.9625*x-1:
d/dx(cu)=c*du/dx where c is a constant.
d/dx(3.9625*x-1)=3.9625*d/dx(x-1)
-The derivative of x-1 is:
d/dx(xn)=nxn-1
d/dx(x-1)=-1*x-1-1
d/dx(x-1)=-1*x-2
d/dx(x-1)=-1/x2
d/dx(3.9625*x-1)=3.9625*(-1/x2)
d/dx(3.9625*x-1)=-3.9625/x2

###### Asked in Math and Arithmetic, Algebra, Calculus

### What is the derivative of 2 to the power of 5x?

25x?
d/dx(au)=au*ln(a)*d/dx(u)
d/dx(25x)=25x*ln(2)*d/dx(5x)
-The derivative of 5x is:
d/dx(cu)=c*du/dx where c is a constant
d/dx(5x)=5*d/dx(x)
d/dx(25x)=95x*ln(2)*(5*d/dx(x))
-The derivative of x is:
d/dx(x)=1x1-1
d/dx(x)=1*x0
d/dx(x)=1*(1)
d/dx(x)=1
d/dx(25x)=25x*ln(2)*(5*1)
d/dx(25x)=25x*ln(2)*(5)
-25x can simplify to (25)x, which equals 32x.
d/dx(95x)=32x*ln(2)*(5)

###### Asked in Math and Arithmetic, Algebra, Calculus

### What is the derivative of ln 1 divided by 1-x?

ln(1)/[1-x]?
d/dx(u/v)=(v*du/dx-u*dv/dx)/(v2)
d/dx(ln(1)/[1-x])=[(1-x)*d/dx(ln1)-ln1*d/dx(1-x)]/[(1-x)2]
-The derivative of ln1 is:
d/dx(lnu)=(1/u)*d/dx(u)
d/dx(ln1)=(1/1)*d/dx(1)
d/dx(ln1)=(1)*d/dx(1)
d/dx(ln1)=d/dx(1)
-The derivative of 1-x is:
d/dx(u-v)=du/dx-dv/dx
d/dx(1-x)=d/dx(1)-d/dx(x)
d/dx(ln(1)/[1-x])=[(1-x)*d/dx(1)-ln1*(d/dx(1)-d/dx(x))]/[(1-x)2]
-The derivative of 1 is 0 because it is a constant.
-The derivative of x is:
d/dx(xn)=nxn-1
d/dx(x)=1*x1-1
d/dx(x)=1*x0
d/dx(x)=1*(1)
d/dx(x)=1
d/dx(ln(1)/[1-x])=[(1-x)*(0)-ln1*(0-1)]/[(1-x)2]
d/dx(ln(1)/[1-x])=[-ln1*(-1)]/[(1-x)2]
d/dx(ln(1)/[1-x])=[ln1]/[(1-x)2]
But you see, ln1 is equal to 0:
d/dx(ln(1)/[1-x])=[0]/[(1-x)2]
d/dx(ln(1)/[1-x])=0

###### Asked in Math and Arithmetic, Algebra, Calculus

### What is the derivative of lnx raised to 4?

ln(x4)?
d/dx(ln(u))=1/u*d/dx(u)
d/dx(ln(x4))=[1/x4]*d/dx(x4)
-The derivative of x4 is:
d/dx(x4)=4x4-1
d/dx(x4)=4x3
d/dx(ln(x4))=[1/x4]*(4x3)
d/dx(ln(x4))=4x3/x4
d/dx(ln(x4))=4/x
(lnx)4?
Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)
d/dx(lnx)4=4(lnx)4-1*d/dx(lnx)
d/dx(lnx)4=4(lnx)3*d/dx(lnx)
-The derivative of lnx is:
d/dx(ln(u))=1/u*d/dx(u)
d/dx(lnx)=1/x*d/dx(x)
d/dx(lnx)=1/x*(1)
d/dx(lnx)=1/x
d/dx(lnx)4=4(lnx)3*(1/x)
d/dx(lnx)4=4(lnx)3/x

###### Asked in Calculus

### What is the derivative of secxtanx?

d/dx(uv)=u*dv/dx+v*du/dx
d/dx(secxtanx)=secx*[d/dx(tanx)]+tanx*[d/dx(secx)]
-The derivative of tanx is:
d/dx(tan u)=[sec(u)]2*d/dx(u)
d/dx(tan x)=[sec(x)]2*d/dx(x)
d/dx(tan x)=[sec(x)]2*(1)
d/dx(tan x)=(sec(x))2=sec2(x)
-The derivative of secx is:
d/dx(sec u)=[sec(u)tan(u)]*d/dx(u)
d/dx(sec x)=[sec(x)tan(x)]*d/dx(x)
d/dx(sec x)=[sec(x)tan(x)]*(1)
d/dx(sec x)=sec(x)tan(x)
d/dx(secxtanx)=secx*[sec2(x)]+tanx*[sec(x)tan(x)]
d/dx(secxtanx)=sec3(x)+sec(x)tan2(x)