Emily reads a story book. o the first day, she reads 1/9 of the whole book, and on the second day, she reads 24 pages. The ratio
2 answers:
There are 270 pages in this book.
Given that,
Emily reads a storybook the first day,
She reads 1/9 of the whole book, and on the second day, she reads 24 pages.
The ratio of the number of pages read to the remaining pages in the two days is 1:4.
We have to find
How many pages are there in this book?
According to the question,
Let, P is the number of pages,
The first day Emily reads p/9 pages of the whole book,
And the second day she read 24 pages.
The ratio of the number of pages read to the remaining pages in the two days = 1;4 = p/5.
Therefore,
The number of pages reads first day + the number of pages read the second day = The ratio of the number of pages read to the remaining pages in the two days
Hence, there are 270 pages in this book.
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then apply the product rule. Hopefully it's clear which function each of f, g, and h refer to.
We then have, using the power and chain rules,
For the third function, we first rewrite in terms of the logarithmic and the exponential functions,
Then by the chain rule,
By the product rule, we have
You could simplify this further if you like.
In Mathematica, you can confirm this by running
D[(4+3x^2)^(1/2) (x^2+1)^(-1/3) Pi^x, x]
The immediate result likely won't match up with what we found earlier, so you could try getting a result that more closely resembles it by following up with Simplify or FullSimplify, as in
FullSimplify[%]
(% refers to the last output)
If it still doesn't match, you can try running
Reduce[<our result> == %, {}]
and if our answer is indeed correct, this will return True. (I don't have access to M at the moment, so I can't check for myself.)
(ii) Given
differentiating both sides with respect to x by the quotient and chain rules, taking y = y(x), gives
which could be simplified further if you wish.
In M, off the top of my head I would suggest verifying this solution by
Solve[D[x*y[x]^3/(1 + Sec[y[x]]) == E^(x*y[x]), x], y'[x]]
but I'm not entirely sure that will work. If you're using version 12 or older (you can check by running $Version), you can use a ResourceFunction,
ResourceFunction["ImplicitD"][<our equation>, x]
but I'm not completely confident that I have the right syntax, so you might want to consult the documentation.