Question

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

$12 \sqrt{x} -x \frac{3}{2} -10$

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:

$y=12 \sqrt{x} -x \frac{3}{2}-10 \\ \\ y=12x^{ \frac{1}{2} } -x \frac{3}{2}-10 \ \ \ \boxed{transform \ \sqrt{x} =x^{ \frac{1}{2} }} \\ \\ Differentiating \\ \\ y'=12. \frac{1}{2}.x^{ \frac{1}{2}-1 } -1.x^{1-1}. \frac{3}{2}-0 \\ \\ y'=6.x^{ -\frac{1}{2} } -x^{0}. \frac{3}{2} \\ \\ y'=6. \frac{1}{x^{ \frac{1}{2} }} -1. \frac{3}{2} \\ \\ \boxed{y'=\frac{6}{ \sqrt{x} } -\frac{3}{2}}$

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

$\frac{6}{ \sqrt{x} } - \frac{3}{2} =0 \\ \\ \frac{6}{ \sqrt{x} } = \frac{3}{2} \\ \\ 3 \sqrt{x} =12 \\ \\ \sqrt{x} = \frac{12}{3} \\ \\ \sqrt{x} =4 \\ \\ (\sqrt{x})^2 =4^2 \\ \\ x=16$