Example: Find dy/dx for the following functions: a) y=x2+1, b) y=x2−1/x2+1, c) y=x/(2x+1)2, d) y=3x+2, e) y=x/(x+4)3, f) y=x2/", g) y=x2/sqrt(x+1), h) y=1/(x+2x-4)2, i) y=sqrt(x−1/2), j) y=-1, k) y=1/(sqrt(x+1))

Acha $u = x^{1/2} - x^{-1}$. Kisha $y = u^2$.
Tafuta $\frac{du}{dx}$:

$$\frac{du}{dx} = \frac{d}{dx}(x^{1/2}) - \frac{d}{dx}(x^{-1})$$ $$\frac{du}{dx} = \frac{1}{2}x^{1/2 - 1} - (-1)x^{-1-1}$$ $$\frac{du}{dx} = \frac{1}{2}x^{-1/2} + x^{-2}$$

Tafuta $\frac{dy}{du}$:

$$\frac{dy}{du} = \frac{d}{du}(u^2) = 2u$$

Badilisha $u$ na $\frac{du}{dx}$:

$$\frac{dy}{dx} = 2\left(x^{1/2} - x^{-1}\right) \left(\frac{1}{2}x^{-1/2} + x^{-2}\right)$$

Panua maneno:

$$\frac{dy}{dx} = 2\left(x^{1/2} \cdot \frac{1}{2}x^{-1/2} + x^{1/2} \cdot x^{-2} - x^{-1} \cdot \frac{1}{2}x^{-1/2} - x^{-1} \cdot x^{-2}\right)$$ $$\frac{dy}{dx} = 2\left(\frac{1}{2}x^{1/2 - 1/2} + x^{1/2 - 2} - \frac{1}{2}x^{-1 - 1/2} - x^{-1 - 2}\right)$$ $$\frac{dy}{dx} = 2\left(\frac{1}{2}x^0 + x^{-3/2} - \frac{1}{2}x^{-3/2} - x^{-3}\right)$$ $$\frac{dy}{dx} = 2\left(\frac{1}{2} + x^{-3/2} - \frac{1}{2}x^{-3/2} - x^{-3}\right)$$

Changanya maneno yanayofanana:

$$\frac{dy}{dx} = 2\left(\frac{1}{2} + \frac{1}{2}x^{-3/2} - x^{-3}\right)$$

Gawanya kwa 2:

$$\frac{dy}{dx} = 1 + x^{-3/2} - 2x^{-3}$$

$$\frac{dy}{dx} = 1 + \frac{1}{x^{3/2}} - \frac{2}{x^3}$$ $$\frac{dy}{dx} = 1 + \frac{1}{x\sqrt{x}} - \frac{2}{x^3}$$

Jibu la mwisho ni:

$$\boxed{\frac{dy}{dx} = 1 + \frac{1}{x\sqrt{x}} - \frac{2}{x^3}}$$