Question

i'am in 11.i'am a medical student i'am havin' a li'll problem in understanding the derivatives in my class... i want 2 solve a numerical using y=uv where y=(5xpower 2 +9) (2xpower 3+4)............. can anyone help me plzzzzzzz.......

Answer 1

I want to define the derivative using $y' = (u \cdot v)'$ , where $y = (5x^2 + 9) \cdot (2x^3 + 4)$ ;

Solution: According to the differentiation product rule for the case of functions $u(x)$ and $v(x)$ :

$(u \cdot v)' = u' \cdot v + u \cdot v'$ , in our case, $u(x) = 5x^2 + 9$ ; $u' = (5x^2 + 9)' = 10x$ ; and $v(x) = 2x^3 + 4$ ; $v' = (2x^3 + 4)' = 6x^2$ ;

Then, $y' = [(5x^2 + 9) \cdot (2x^3 + 4)]' = 10x \cdot (2x^3 + 4) + (5x^2 + 9) \cdot 6x^2 = 20x^4 + 40x + 30x^4 + 54x^2 = 50x^4 + 54x^2 + 40x$ .

Answer: $y' = 50x^4 + 54x^2 + 40x$ .