Question

I found this question in an old exam paper, but not sure how to handle it.

Answer 1

I'm supposed to find the derivative of h.

Answer 2

I think you should integrate the expression first?

Answer 3

I'm still learning integration so I'm not sure how to handle that yet. The top is not quite the derivative of the bottom or it would've been easy.

Answer 4

It's actually the derivative of it. :))
Do you remember that\[\int\limits_{}^{} \frac{ dz }{ z^2+1 }\] is actually equal to tangent x so letting our z^4 = z^2
Therefore, z = z^2 and derivative of z^2 is... :)))

Answer 5

You lost me...

Answer 6

using the definintion of integrations, let:
\[F(x)=\int_{a(x)}^{b(x)}f(t)dt\]
therefore
\[F(x)=F[b(x)]-F[a(x)]\]
taking the derivative, and applying the chain rule gives us:
\[\frac d{dx}F(x)=f[b(x)]~b'-f[a(x)]~a'\]

Answer 7

in other words, integrating it is not a necessary step. not wrong, just not needed.

Answer 8

\[h(x)=-\int_{\sqrt x}^{1}\frac{2z^2}{z^4+1}dz\]
\[\frac d{dx}h(x)=-\frac{d}{dx}~\int_{\sqrt x}^{1}\frac{2z^2}{z^4+1}dz\]
\[\frac d{dx}h(x)=-\left(\frac{2(1)^2}{(1)^4+1}(1)'-\frac{2(\sqrt x)^2}{(\sqrt x)^4+1}~(\sqrt x)'\right)\]

Answer 9

to reiterate; but this time lets write it as f'(t) to indicate its true nature as the derivative of some function that we will integrate back into
\[F(x)=\int_{a(x)}^{b(x)}f'(t)dt\]
\[\frac{d}{dx}F(x)=\frac{d}{dx}\int_{a(x)}^{b(x)}f'(t)dt\]
notice that we are just taking the derivative of the function that we are integrating up into - so we will get back to this f'(t):
\[\frac{d}{dx}F(x)=\frac{d}{dx}(f[b(x)]-f[a(x)])\]
\[\frac{d}{dx}F(x)=\frac{d}{dx}f[b(x)]-\frac{d}{dx}f[a(x)]\]

now by taking the derivative of the sum ... and applying the chain rule
\[\frac{d}{dx}F(x)=f'[b(x)]~b'-f'[a(x)]~a'\]
and f' was already defined for us as the function getting integrated to start with.

Answer 10

I'm terribly confused but I'll mull over it when I get home. Thanks though. I have more to learn about integration than I realized.

Answer 11

:) its just applying the definition of integration and derivative together is all.

Answer 12

to simplify the long drawn out version, given:
\[F(x)=\int_{a}^{b}f(t)~dt\]the derivative is:
\[\frac{d}{dx}F(x)=f[b]b'-f[a]a'\]