Question

What is the derivative of this function
and what is the normal slope at x=5

Answer 1

\[x-\sqrt{(x-1)}\]

Answer 2

$$f(x)=x-\sqrt{x-1}\\f'(x)=1-\frac1{2\sqrt{x-1}}$$

Answer 3

I keep trying it but i don't get the answer, I'm using the product rule

Answer 4

Now recall that the normal slope is that of the line perpendicular to our tangent...

Answer 5

Just use chain rule:$$\frac{d\sqrt{x-1}}{dx}=\frac{d\sqrt{x-1}}{d(x-1)}\times\frac{d(x-1)}{dx}=\frac1{2\sqrt{x-1}}\times1=\frac1{2\sqrt{x-1}}$$

Answer 6

i was using both

Answer 7

@Zarkon please help