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[(- 3xy) square times x to the 3rd power - 2x square times (3xy Square) cube times Y / 2] divided by 9x to the 4th power y to the 3rd power Followed by y to the power of 2
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Given xy = 9 and X-Y = - 3, find the value of the square of the third power of X + 3xy + y
x^3+3xy+y^3
=（x^3+y^3）+3xy
=(x+y)(x^2-xy+y^2)+3xy
x^2+y^2=(x-y)^2+2xy
=9+18=27
So (x + y) (x ^ 2-xy + y ^ 2) + 3xy
=(-3)*(27-9)+3*9
=-54+27
=-27
Your question may be wrong. It should be the cube of Y
① (- 1 / 2x square y) to the third power (- 3xy Square) ② Square of (- x) * cube of X (- 2Y) cube of + (2XY) square of (- x) cube y
(-1/2x ² y) ³ (-3xy ²)²
=(- 1 / 8x to the 6th power of Y) ³) (9x ² Y to the fourth power)
=(-1/8 × 9) The (6 + 2) power of X and the (3 + 4) power of Y
=-The 8th power of 9 / 8x and the 7th power of Y
(-x) ²* x ³ (-2y) ³+ (2xy) ² (-x) ³ y
=x ²* x ³* (-8y ³)+ (4x ² y ²) (-x ³) y
=-8x to the 5th power y ³- 4X to the 5th power y ³
=-12x to the 5th power y ³
The third power of (- 2x) multiplied by the square of (- 3xy)
（-3x)^3*(-3xy^2)
=-27x^3*(-3xy^2)
=81x^(3+1)*y^2
=81x^4y^2
Find the differential of [x ^ 2 * (SiNx) ^ 2] / (1 + x ^ 2)
y=[x^2*(sinx)^2]/(1+x^2)
lny=lnx^2+ln(sinx)^2-ln(1+x^2)
y'/y=2/x+2cotx-2x/(1+x^2)
y‘=y[2/x+2cotx-2x/(1+x^2)]
=[x^2*(sinx)^2]/(1+x^2)*[2/x+2cotx-2x/(1+x^2)]
dy=
[x^2*(sinx)^2]/(1+x^2)*[2/x+2cotx-2x/(1+x^2)]dx
Proving that LNX is approximately equal to X by differentiation
error
It should be when | x | < 1, ln (1 + x) ≈ X
f(x)≈f(x0)+f'(x0) Δ x
Note x0 = 1, Δ x=x
f'(x)=1/x,
ln(1+x)≈ln1+x=x
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