Q.Find the derivative of

(i) $2x - \dfrac{3}{4}$

Key idea: Term-by-term differentiation using the power rule and constant rule.

• Derivative of $2x$ is $2$. …

1. (i) $2x - \dfrac{3}{4}$

   This is a simple linear function. The derivative of $2x$ is $2$ (since $\frac{d}{dx}(x) = 1$). The constant term $-\frac{3}{4}$ differentiates to $0$.

   So, $\frac{d}{dx}\left(2x - \frac{3}{4}\right) = 2$.

2. (ii) $(5x^3 + 3x - 1)(x - 1)$

   Here we have a product of two functions. Let $u = 5x^3 + 3x - 1$ and $v = x - 1$.

   Using the product rule: $\frac{d}{dx}(uv) = u'v + uv'$.

   First, find $u' = 15x^2 + 3$ and $v' = 1$.

   Then,

Adding them:  

So, the derivative is $20x^3 - 15x^2 + 6x - 4$.

3. (iii) $x^{-3}(5 + 3x)$

Rewrite as $5x^{-3} + 3x^{-2}$. Now apply the power rule: $\frac{d}{dx}(x^n) = nx^{n-1}$.

For $5x^{-3}$: $5 \cdot (-3)x^{-4} = -15x^{-4}$.

For $3x^{-2}$: $3 \cdot (-2)x^{-3} = -6x^{-3}$.

So, the derivative is $-15x^{-4} - 6x^{-3}$.

4. (iv) $x^5(3 - 6x^{-9})$

   Expand first: $3x^5 - 6x^{-4}$.

   Differentiate term by term:

   For $3x^5$: $3 \cdot 5x^4 = 15x^4$.

   For $-6x^{-4}$: $-6 \cdot (-4)x^{-5} = 24x^{-5}$.

   So, the derivative is $15x^4 + 24x^{-5}$.

5. (v) $x^{-4}(3 - 4x^{-5})$

   Expand: $3x^{-4} - 4x^{-9}$.

   Differentiate:

   For $3x^{-4}$: $3 \cdot (-4)x^{-5} = -12x^{-5}$.

   For $-4x^{-9}$: $-4 \cdot (-9)x^{-10} = 36x^{-10}$.

   So, the derivative is $-12x^{-5} + 36x^{-10}$.

6. (vi) $\dfrac{2}{x + 1} - \dfrac{x^2}{3x - 1}$

   This requires the quotient rule for each term. The quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.

   First term: $\frac{2}{x+1}$. Here $u=2$, $v=x+1$.

   $u'=0$, $v'=1$.

   Derivative = $\frac{0 \cdot (x+1) - 2 \cdot 1}{(x+1)^2} = \frac{-2}{(x+1)^2}$.

   Second term: $\frac{x^2}{3x-1}$. Here $u=x^2$, $v=3x-1$. …