Question

For some constants, a and b, find the derivative of
(i) (x − a) (x − b)
Please do not copy

Answer 1

$\sf\large\underline{\underline{\red{\sf Question:}}}$

For some constants, a and b, find the derivative of

• (i) (x − a) (x − b)  

$\sf\large\underline{\underline{\red{\sf Solution:}}}$

$\purple{\underbrace{\sf \pink{Method\;1}}}:$

Let f(x) = (x-a)(x-b)

Simplifying :

f(x) = x(x-b)-a(x-b)

f(x) = x² -bx -ax + b

differentiating with respect to x,

$\sf \red\bigstar\boxed{\sf\dfrac{d}{dx} a^n= n(a)^{n-1}}$

f'(x) = 2x -b -a + 0

So,

f'(x) = 2x -b-a = 2x - 1(b+a)

$\purple{\underbrace{\sf \pink{Method\;2}}}:$

Using product formula,

$\sf \red\bigstar$ (UV)' = U(V)' + V(U)'

Given to find derivative of (x − a) (x − b)

Here we have ,

• U = (x − a)
• V (x − b)

f(x) = (x − a) (x − b)

f'(x) = (x − a) (x − b)' + (x − b) (x − a)'

f'(x) = (x- a) (1) + (x-b)(1)

f'(x) = (x-a) +(x-b)

f'(x) = x - a + x -b

f'(x) = 2x -a -b = 2x -1(a +b)

Henceforth,

• Derivative of (x-a)(x-b) is 2x -1(a +b).

$\sf\large\underline{\underline{\red{\sf More\;to\;know\; :}}}$

• $\bf\left(\dfrac{U}{V}\right)' = \dfrac{V(U)' -U(V)'}{V^2}$
• derivative of x w.r.t x is always 1.
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• derivative of constant term is zero(0).
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• (cos x )' = - sin x
• ( tan x )' = sec² x
• (sec x)' = sec x.tan x