Fundamental theorem of Calculus: find the derivative...

[confused_07]

Fundamental theorem of Calculus: find the derivative of:
h[x]= int [x^2,2] sqrt(u-1) du

Here's what I did:

Let y= h[x] and t= x^2

y= int [t,2] (u-1)^(1/2) du
dy/dt= (t-1)^(1/2)
= [1/(2sqrtx)] * (t-1) * 1

Chain Rule:
f[t]= t^(1/2)
f'[t]= (1/2)t^(-1/2)
= 1/(2sqrt(t))
= 1/(2x)
g[t]= t-1
g'[t]= 2x

h'[x]= [1/(2x)] (x^2-1)*(2x)
= [(x^2/2x)-(1/2x)](2x)
= 2x(x^2/2x)
= x^2

This does not look right... where did I go wrong ( I am sure it's in the substitution)?

 

[galactus]

I am having a rough time understanding your notation.

Why don't you consider learning a little LaTex?. It would make things so much readable if you plan on posting to any extent.

The Second Fundamental Theorem of Calculus?. Is this what you mean:

\(\displaystyle \L\\\frac{d}{dx}\int_{2}^{x^{2}}\sqrt{u-1}du\)

 

[confused_07]

Yes, apologize..... I am having a hardtime learning calc, let alone Tex

 

[pka]

\(\displaystyle \L \frac{d}{{dx}}\left[ {\int\limits_2^{\quad x^2 } {\sqrt {u - 1} du} } \right] = 2x\sqrt {x^2 - 1}.\)

 

[confused_07]

Wow... was I off. I was following my text book, which is why I am confused. Can you explain how you came to that answer. I do better when I see the steps laid out, then I usually figure it out. I am sure it's a lot of typing, but I REALLY do appreciate it.

Thank you by the way......

 

[pka]

Suppose that g & h are differentiable functions then:

\(\displaystyle \L \frac{d}{{dx}}\left[ {\int\limits_{g(x)}^{h(x)} {f(u)du} } \right] = h'(x)f(h(x)) - g'(x)f(g(x))\).