Question

If y=(x²-2x+3)(x²-3x+15), Find dy/dx by (1) Product rule (2) Multiply and using rule for polynomials. (3) Logatithmic differentation and compare.

Answer 1

Answer:

In all cases derivative is same

Step-by-step explanation:

Using product rule first

Since we are given that

y = (x²-2x+3)(x²-3x+15)

Taking derivatives on both sides we get

dy/dx = d((x²-2x+3)(x²-3x+15))/dx

Using the product rule we get

dy/dx = (x²-2x+3)d((x²-3x+15))/dx + (d(x²-2x+3)/dx)(x²-3x+15)

          = (x²-2x+3)(2x - 3) + ( 2x - 2)(x²-3x+15)

          = 2x³ - 3x² - 4x² + 6x + 6x - 9 + 2x³ - 6x² + 30x - 2x² + 6x - 30

          = 4x³ - 15x² + 48x - 39

So

dy/dx = 4x³ - 15x² + 48x - 39      ....(1)

NOW

Multiplying first then using polynomial rule

Since

y = (x² - 2x + 3)(x²- 3x + 15)

By multiplying polynomials we get

y = x^4 - 3x³ + 15x² - 2x³ + 6x² - 30x + 3x² - 9x + 45

Taking derivative on both sides we get

dy/dx = 4x³ - 9x² + 30x - 6x² + 12x - 30 +  6x - 9

          = 4x³ - 15x² + 48x - 39  

Thus

dy/dx = 4x³ - 15x² + 48x - 39     .....(2)

From equation (1) and (2) it is clear the on both case derivative is same