My Twelfth Problem

Calculus Level 3

tan ( ln ( x + y ) ) = sinh 10 ( x + y ) \large \tan (\ln(x+y))=\sinh^{10}(x+y)

Given the above, evaluate d y d x \dfrac{dy}{dx} for x = π 6 x=\dfrac{\pi}{6} and 1 < y < 0 -1<y<0 .


The answer is -1.

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1 solution

Daniel Ellesar
Nov 2, 2015

Any x to y graph that takes the form f ( x + y ) = 0 f(x+y)=0 is linear with a constant gradient of 1 -1 (unless it has weird things like mods that could change the x + y x+y bit), where the axis intercepts are the solution(s) for z z to the equation f ( z ) = 0 f(z)=0 . A graph of this particular equation takes the form of infinite parallel lines, each with a gradient of 1 -1 , where the axis intercepts are the solutions for z z to the equation t a n [ l n ( z ) ] s i n h 10 ( z ) = 0 tan[ln(z)]-sinh^{10}(z)=0 (which I can't do in my head).

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