What's dy/dx of math?

Hi Christopher !!!!!!!!!!!!!!!!!!!!!!!!!!!!!


Here you go sweetheart :)

1) i) y = (1-x) /x²

After simplifying this equation we get :

y = x^(-2) - x^(-1)

Differentiating gives us :

dy/dx = - 2/x³ + 1/x²

We want dy/dx = 0

So setting this equal to zero and solving for 'x' we have :

dy/dx = 0
- 2/x³ + 1/x² = 0
2/x³ = 1/x²
2x² = x³
x³ - 2x² = 0
x² (x - 2) = 0

So x = 0 or x = 2

Since x = 0 is not defined for this function, (it is a vertical asymptote), we reject x = 0 and accept that at x = 2 this is a stationary point (a minimum value for the curve)


ii) y=(1+x)(1-x)²

After simplifying this equation we have :

y = x³ - x² - x + 1

Differentiating this gives us :

dy/dx = 3x² - 2x - 1

We want dy/dx = 0,

So set dy/dx = 0 and so solve for 'x' :

dy/dx =0
3x² - 2x - 1 = 0
(3x+ 1)(x - 1) = 0
x = - 1/3 or x = 1

So the two values of 'x' which have gradient zero are x = -1/3 and x = 1.


2) Assuming this is what YOU mean :

θ=π+(x sin θ), find dx/dθ.

We do this by differentiating implicitly and using the product rule :

∴ dθ/dx = 0 + [xcosθ + dθ/dx(sinθ)]

Now we rearrange and solve for dθ/dx :

dθ/dx - (dθ/dx sinθ) = xcosθ

dθ/dx (1 - sinθ) = xcosθ

dθ/dx = xcosθ /(1 - sinθ)


But we WANT dx/dθ, so take the reciprocal :

∴ dx/dθ = (1 - sinθ) /xcosθ



Hope this helps !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
P.S. (Don't forget to vote me best answer as being the first to correctly and thoroughly answer your question!)

1.
(i)
y = (1 - x)/x²

dy/dx = (d/dx)[(1 - x)/x²]
dy/dx = [x²•(d/dx)(1 - x) - (1 - x)•(d/dx)x²] / x⁴
dy/dx = [x²•(-1) - (1 - x)•(2x)] / x⁴
dy/dx = (-x² - 2x + 2x²) / x⁴
dy/dx = (x² - 2x) / x⁴
dy/dx = (x - 2) / x³

When gradient = 0 :
(x - 2) / x³ = 0
x - 2 = 0 and x ≠ 0
x = 2


(ii)
y = (1 + x)(1 - x)²

dy/dx = (d/dx)[(1 + x)(1 - x)²]
dy/dx = (1 + x)•(d/dx)(1 - x)² + (1 - x)²•(d/dx)(1 + x)
dy/dx = (1 + x)•2(1 - x)•(-1) + (1 - x)²•(1)
dy/dx = -2(1 + x)(1 - x) + (1 - x)²
dy/dx = (1 - x)[-2(1 + x) + 1 - x]
dy/dx = (1 - x)(-2 - 2x + 1 - x)
dy/dx = (x - 1)(3x + 1)

When gradient = 0 :
(x - 1)(3x + 1) = 0
x - 1 = 0 or 3x + 1 = 0
x = 1 or x = -1/3


====
2.
θ = π + x sinθ
(d/dθ)θ = (d/dθ)(π + x sinθ)
1 = (d/dθ)π + (d/dθ)(x sinθ)
0 + x•(d/dθ)sinθ + sinθ•(d/dθ)x = 1
x cosθ + sinθ (dx/dθ) = 1
sinθ (dx/dθ) = 1 - x cosθ
dx/dθ = (1 - x cosθ) / sinθ
dx/dθ = cscθ - x cotθ

1.
i)
y=(1-x)/x^2

apply the quotient rule
u(x)=1-x
u'(x)=-1
v(x)=x^2
v'(x)=2x

dy/dx = (v(x) u'(x) - u(x) v'(x) ) / v^2(x)
dy/dx = (x^2 (-1) - (1-x) (2x) ) / x^4
dy/dx = (-x^2-2x+2x^2)/x^4
dy/dx = (x^2-2x) /x^4
dy/dx = x(x-2)/x^4
= (x-2)/x^3

ii)
y= (1+x)(1-x)^2

apply the product rule
u(x) = 1+x
u'(x) = 1
v(x) = (1-x)^2
v'(x) = 2(1-x)(-1) = -2(1-x) = -2+2x

dy/dx = u(x)v'(x) + v(x) u'(x)
dy/dx = (1+x) (-2+2x) + (1-x)^2 (1)
dy/dx = -2+2x-2x+2x^2 + 1-2x+x^2
dy/dx = 3x^2-2x-1

2.
θ = π + x sin θ
x sin θ = θ - π
x = (θ-PI) / sin θ
x = (θ-π) csc θ

dx/dθ = (θ-π) d/dθ (csc θ) + csc θ d/dθ(θ-π)
dx/dθ = (θ-π) (-csc θ cot θ) + csc θ (1)
dx/dθ = -θ csc θ cot θ + π csc θ cot θ + csc θ

1.
(i) y = (1 - x)/x²
y' = (-1 * x^2 - 2x * (1 - x))/(x^2)^2
y' = (x^2 - 2x)/x^4
y' = (x - 2)/x^3

y' = 0
(x^2 - 2x)/x^4 = 0
x = 2

(ii) y = (1 + x)(1 - x)²
y = x^3 - x^2 - x + 1
y' = 3x^2 - 2x - 1

y' = 0
3x^2 - 2x - 1 = 0
(x - 1)(3x + 1) = 0
x = -1/3, 1


2. θ = π + x sin θ
x = (θ - π)/sin θ

dx/dθ = (1 * sin θ - cos θ * (θ - π)) / sin² θ