Equation for Tangent at (1,2) of y=2(sqrt(x)) | Derivatives Tutorial

[ur5pointos2sl]

Find an equation for the tangent to the curve at the given point.

y=2(sqrt(x)) point- (1,2)

Ok so I think I can work most of it then I get stuck. I am not sure how you guys use the codes so I will try to type it out the best I can.

lim = ( f(a+h)-f(a) ) / (h)
h->0
= f(1+h)-f(1) / h
= (2 sqrt(1+h) - 2) / h

This is where I have gotten to. substituting in now will give me a 0 in my denominator. So where would I go next?

 

[danago]

Using the limit definition of the derivative is often quite tedious to actually find a derivative. Have you seen the result:

\frac{d}{dx} x^n=nx^{n-1}

That will be much more useful for finding the derivative of your function.

 

[ur5pointos2sl]

Using the limit definition of the derivative is often quite tedious to actually find a derivative. Have you seen the result:

\frac{d}{dx} x^n=nx^{n-1}

That will be much more useful for finding the derivative of your function.

The teacher has not yet introduced us to that form to find the derivatives yet. He says that's the easy way. he just gave us the formula f(a+h)-f(a) / h and told us to use that until we actually get into derivatives.

The answer is y = x+1