Differentiation with respect to x' of the following function:
(1) (x2-5x+6) (x-3)
did you mean
y = (x^2 - 5x+6)(x-3) ?
For this one, I would have actually expanded it.
y = x^3 - 3x^2 - 5x^2 + 15x + 6x - 18
= x^3 - 8x^2 + 21x - 18
dy/dx = 3x^2 - 16x + 21
or .... by the product rule:
dy/dx = (x^2 - 5x + 6)(1) + (2x - 5)(x-3)
= x^2 - 5x + 6 + 2x^2 - 6x - 5x + 15
= 3x^2 - 16x + 21
Thankyou sir
To differentiate the given function (1) with respect to x', we need to consider x' as a function of x. Let's denote x' as f(x).
First, we need to expand the function (1) using the distributive property:
(1) (x^2 - 5x + 6) (x - 3)
Expanding this equation gives:
x^3 - 3x^2 - 5x^2 + 15x + 6x - 18
Simplifying further, we have:
x^3 - 8x^2 + 21x - 18
Now, we differentiate the equation with respect to the variable x:
d/dx (x^3 - 8x^2 + 21x - 18)
This gives:
3x^2 - 16x + 21
Finally, we substitute x' back into the equation by replacing x with f(x):
3f(x)^2 - 16f(x) + 21
Therefore, the differentiation with respect to x' of the function (1) is given by:
3f(x)^2 - 16f(x) + 21, where x' = f(x).