Calculate the slope of the tangent to each curve at the given value of x using first principles.
f(x)= (x+4)^1/2 , x=5

Question

Calculate the slope of the tangent to each curve at the given value of x using first principles. 
f(x)= (x+4)^1/2     , x=5

anonymous · Answer

first principles?
The easiest way to do this is with calculus, by taking the derivative of f(x).
$$f'(x)=1/2(x+4){^{-1/2}}$$
and then plugging in the x=5, so f'(x)=-1/6 and 1/6

anonymous · Answer

Thank you for responding to my question. I know how to do the question using derivates but I can't quite remember how to do it with first principles because we'll be asked to do so on the exam.

anonymous · Answer

Hey, so I'm not sure what first principles are. Could you explain those?

anonymous · Answer

m= f(x+h) - f(x) / h  basically the harder way to find the slope than using derivatives. lol

anonymous · Answer

Ok I think I got them
So m= f(x+h) - f(x) / h
So then the slope is
$$m=Deltay/Deltax=dy/dx$$
So
$$dy/dx=f(x+h)-f(x)/h$$
Thus, $$dy/dx=\lim_{h \rightarrow 0} f(x+h)-f(x)/h$$\]
So since f(x)=(x+4)^1/2, f(x+h)=(x+4+h)^1/2
So the equation is
$$m=(x+4+h) ^{1/2}-(x+4)^{1/2}/h$$
$$dy/dx=\lim_{h \rightarrow 0}(x+4+h) ^{1/2}-(x+4)^{1/2}/h$$

anonymous · Answer

oh right okay, thank you!

anonymous · Answer

I don't know how to take the limit of that, but at least it's a start :)

anonymous · Answer

yep I know where to go from there so thanks :)