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Question
Differentiate the following:
h(t) = `("t" - 1/"t")^(3/2)`
Solution
h(t) = `("t" - 1/"t")^(3/2)`
[y = f(g(x)
`("d"y)/("d"x)` = f'(g(x)) . g'(x)]
h'(t) = `3/2 ("t" - 1/"t")^(3/2 - 1) "d"/"dt" ("t" - 1/"t")`
= `3/2 (1 - 1/"t")^(1/2) [ 1 - (- 1) "t"^(-1 - 1)]`
= `3/2 ("t" - 1/"t")^(1/2) (1 + "t"^(-2))`
h'(t) = `3/2 ("t" - 1/"t")^(1/2) (1 + 1/"t"^2)`
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