Question

2. Find derivatives of the following functions by applying the product rule: (i) y = (3-2)(x-x+1)
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Answer 1

Answer:

derivative of above function is 0.

Explanation:

As, x will cancel the other x.

then, y = 1

y will be a constant and derivative of a constant is 0.

Answer 2

Given,

y = (3 - x)(x - x + 1)

Further simplifying y, we get, y = (3 - x)(1)

AND, Obtained expression can be differentiated much easily without applying any (Product) Rule.

$\sf \frac{d(y)}{dx} = \frac{d(3 - x)}{dx} = 3 - 0 = \green3$

But,

• $\text{ If the expression had been} \: \\ \sf y = (3 - x)(x^2 - x + 1)$

Now, THE PRODUCT RULE COULD HAVE BEEN APPLIED

So,

$\sf \frac{d(y)}{dx} \\ \\ \sf= \frac{d}{dx} \bigg\{ (3 - x)( {x}^{2} - x + 1 )\bigg\} \\ \\\small \sf = \bigg[(3 - x) \frac{d}{dx} \{ {x}^{2} - x + 1\} \bigg]+ \bigg[( {x}^{2} - x + 1) \frac{d}{dx} \{3 - x\} \bigg] \\ \\\sf = \bigg[(3 - x)(2x - 1)\bigg] +\bigg[( {x}^{2} - x+ 1 )( - 1) \bigg] \\ \\ \sf =6x - {2x}^{2} - 3 + x + ( - {x}^{2} + x - 1) \\ \sf =7x - {x}^{2} - 3 - {x}^{2} + x - 1 \\ \sf = \red{8x - {2x}^{2} - 4}$

• Hope any of the above solutions help you out☺