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Differentiation of an integral

Posted 8 years ago

I am new to Mathematica. i hope to have the software next week so please excuse the unusual syntax.

p(y) means p is a function of y. u(p) means u is a function of p p(x) is the value of p when y=x

I need to differentiate wrt y the following: p(x)*integral between x and h((1/p(y) * u(p) * dp/dy * dy) Differentiating by parts i get
u * dp/dy + dp/dyintegral between x and h((1/p) * u * dp/dy dy)

Is this correct?

POSTED BY: John Mc Kiernan
4 Replies

The derivative of the whole thing is still 0. For

$p(y) = a + b y$

we have

$p'(y) = b$

so the integrand

$1/p(y) * u(p) * p'(y)$

collapses to

$\frac{b u(y)}{a + b y}$.

Thus the entire expression whose derivative you want is:

$p(x) \int_x^h \frac{b u(y)}{a + b y} dy$.

The derivative of that with respect to $y$ — which is what you asked for — is still 0.

Or did you really want the derivative with respect to $x$??

POSTED BY: Murray Eisenberg

P is also a function of y ,p(y)=a+by where a and b are constants.

POSTED BY: John Mc Kiernan

Apologies u is a function of p.i.e u(p).I have corrected (edited) the script in the original question

POSTED BY: John Mc Kiernan

This is the same question you posted on mathematica.stackexchange, and with the same syntax issue:

 Integral between x and h((1/p(y) * u(y) * dp/dy * dy)

Do you mean the integral of (1/p(y) * u(y) * dp/dy * dy from x to h?

If so, note that the value of the integral does not depend upon the dummy variable y, so its derivative with respect to y will be zero.

POSTED BY: Murray Eisenberg
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