Locate the turning points on the curve y=3x2−6x and 2. Determine dx/dy and dz/dy.

Kazi uliyopewa ni $z = \frac{1}{\sqrt{x^2 + y^2}}$.
Hii inaweza kuandikwa kama:

$$z = (x^2 + y^2)^{-1/2}$$

Ili kupata $\frac{dz}{dx}$, tunachukulia $y$ kama kigezo kisichobadilika (constant). Tutatumia chain rule.
Kanuni ya chain rule ni $\frac{d}{dx}[u^n] = n u^{n-1} \frac{du}{dx}$.
Hapa, $u = x^2 + y^2$ na $n = -\frac{1}{2}$.
Kwanza, pata $\frac{du}{dx}$:

$$\frac{du}{dx} = \frac{d}{dx}(x^2 + y^2) = 2x + 0 = 2x$$

Sasa, tumia chain rule:

$$\frac{dz}{dx} = -\frac{1}{2}(x^2 + y^2)^{-1/2 - 1} \cdot (2x)$$ $$\frac{dz}{dx} = -\frac{1}{2}(x^2 + y^2)^{-3/2} \cdot (2x)$$

Rahisisha usemi:

$$\frac{dz}{dx} = -x(x^2 + y^2)^{-3/2}$$

Au kwa fomu ya mizizi:

$$\frac{dz}{dx} = -\frac{x}{(x^2 + y^2)^{3/2}}$$

Ili kupata $\frac{dz}{dy}$, tunachukulia $x$ kama kigezo kisichobadilika (constant). Tutatumia chain rule tena.
Hapa, $u = x^2 + y^2$ na $n = -\frac{1}{2}$.
Kwanza, pata $\frac{du}{dy}$:

$$\frac{du}{dy} = \frac{d}{dy}(x^2 + y^2) = 0 + 2y = 2y$$

Sasa, tumia chain rule:

$$\frac{dz}{dy} = -\frac{1}{2}(x^2 + y^2)^{-1/2 - 1} \cdot (2y)$$ $$\frac{dz}{dy} = -\frac{1}{2}(x^2 + y^2)^{-3/2} \cdot (2y)$$

Rahisisha usemi:

$$\frac{dz}{dy} = -y(x^2 + y^2)^{-3/2}$$

Au kwa fomu ya mizizi:

$$\frac{dz}{dy} = -\frac{y}{(x^2 + y^2)^{3/2}}$$

Majibu ni:

$$\boxed{\frac{dz}{dx} = -\frac{x}{(x^2 + y^2)^{3/2}}}$$ $$\boxed{\frac{dz}{dy} = -\frac{y}{(x^2 + y^2)^{3/2}}}$$