Hello all, just after a bit of help with this question, do i show this by taking the derivative? i'm not sure.. Show that for all
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As you can see above, and . The hypotenuse then becomes (Pythagoras' theorem).
Hello, Robb! No derivatives needed . . . It's an Identity problem. Show that: . . for all .[1] Then we have: . is in a right triangle with: Using Pythagorus, we have: . Hence: . .[2] Equate [1] and [2]: .
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