Use CALCULUS to find coordinates of the turning point on C. 12 √(x) -x (3)/(2) -10 I know I have differentiate etc., but I'm struggling with the differentiation! This is AS…

Question

Use CALCULUS to find coordinates of the turning point on C.


`12 √(x) -x (3)/(2) -10`

I know I have differentiate etc., but I'm struggling with the differentiation!
This is AS maths, Core 1.

Thanks! :)

Answer 1

Oi

Just calculating the differential. Step by step:


`y=12 √(x) -x (3)/(2)-10 \\ \\ y=12x^{ (1)/(2) } -x (3)/(2)-10 \ \ \ \boxed{transform \ √(x) =x^{ (1)/(2) }} \\ \\ Differentiating \\ \\ y'=12. (1)/(2).x^{ (1)/(2)-1 } -1.x^(1-1). (3)/(2)-0 \\ \\ y'=6.x^{ -(1)/(2) } -x^(0). (3)/(2) \\ \\ y'=6. \frac{1}{x^{ (1)/(2) }} -1. (3)/(2) \\ \\ \boxed{y'=(6)/( √(x) ) -(3)/(2)}`

If you want to know where the tangent has no slope:


`(6)/( √(x) ) - (3)/(2) =0 \\ \\ (6)/( √(x) ) = (3)/(2) \\ \\ 3 √(x) =12 \\ \\ √(x) = (12)/(3) \\ \\ √(x) =4 \\ \\ (√(x))^2 =4^2 \\ \\ x=16`