$35pq^2=5\cdot7\cdot p\cdot q^2\\-14pq^2=-1 \cdot 2\cdot 7\cdot p \cdot q^2\\-21p^2q^3=-1\cdot 3\cdot 7\cdot p^2 \cdot q^3\\\\\text{hcf}(35pq^2,-14pq^2,-21p^2q^3)=7\cdot p \cdot q^2=7pq^2$

6 0

3 years ago

3√x -2/x^2<br><br>please show step by step of differentiation before combining the terms.

EastWind [94]

Answer:

Step-by-step explanation:

Before we differentiate, let us assign a variable to the function. Let y be equal to the function i.e let y = 3√x -2/x²

In differentiation if $y = ax^{n}$ , then $\frac{dy}{dx} = nax^{n-1}$ where n is a constant and dy/dx means we are differentiating the function y with respect to x.

Applying the formula o the question given;

$y= 3\sqrt{x} -2/x^2\\y = 3{x}^\frac{1}{2} - 2x^{-2} \\\\$

On differentiating the resulting function;

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