Question

y=(x-1)(x-2)(x-3)- - - - -(x-2020). find its 1st ordered derivative

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Answer 1

Question:

Find the 1st ordered derivative of ,

y = (x-1)(x-2)(x-3)-----(x-2020) .

Solution:

Please refer to the attachment.

Note:

1. Sum rule and difference rule:

If y = f(x) ± g(x) , then ;

dy/dx = d{f(x)}/dx ± d{g(x)}/dx

2. Product rule:

If y = f(x)•g(x) , then ;

dy/dx = g(x)•d{f(x)}/dx + f(x)•d{g(x)}/dx

3. Division rule:

If y = f(x)/g(x) , then ;

dy/dx = [g(x)•d{f(x)}/dx - f(x)•d{g(x)}/dx] / {g(x)}²

4. Chain rule:

If y = f(u) , u = g(v) , v = h(x) , then ;

dy/dx = (dy/du)•(du/dv)•(dv/dx)

• d(logx)/dx = 1/x

• log(A•B•C•...) = logA + logB + logC +...

• log(A/B) = logA - logB

• log(A^B) = B•logA

Answer 2

Answer:-

REQUIRED TO FIND :-

• First order derivative of y = (x-1) (x-2) (x-3) ......(x-2020)

SOLUTION :-

$Given \: \: \\ \\ y = (x-1) (x-2) (x-3) ......(x-2020) \\ \\ APPLYING \: \: LOG \: \: ON \: \: BOTH \: \: SIDES \\ \\ log(y) = log(x-1) \: + log(x - 2) \: + log(x-3) \: + .......log(x - 2020) \: + \\ \\ DIFFERENTIATING \: \: \: W.R.T \: \: X \: \: ON \: \: BOTH SIDES \\ \\ \frac{1}{y} \frac{dy}{dx} \: = \frac{1}{x - 1} \: + \frac{1}{x - 2} \: +\frac{1}{x - 3} \: +........\frac{1}{x - 2020} \\ \\ \frac{dy}{dx} \: = y(\frac{1}{x - 1} \: + \frac{1}{x - 2} \: +\frac{1}{x - 3} \: +........\frac{1}{x - 2020} ) \\ \\ \frac{dy}{dx} = (x-1) (x-2) (x-3) ......(x-2020)( \: \frac{1}{x - 1} \: + \frac{1}{x - 2} \: +\frac{1}{x - 3} \: +........\frac{1}{x - 2020} )$

NOTE :-

$\frac{d log(x) }{dx} \: = \frac{1}{x} \\ \\ log(ab) = log(a) \: + \: log(b)$

(uvw)' = u'vw + uv'w + uvw'

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