Trig Implicit Derivative

[Jason76]

Post Edited

Solving for \(\displaystyle y'\)

\(\displaystyle f(x) = 9\cos(x) \sin(y)= 7\) - Use product rule on left \(\displaystyle g(f') + f(g')\) once doing derivative.

\(\displaystyle [\sin(y)][-9\sin(x)] + [9\cos(x)][\cos(y)]y' = 0\)

\(\displaystyle -9\sin(y) \sin(x) + 9\cos(x)\cos(y)y' = 0\)

\(\displaystyle 9\cos(x)\cos(y)y' = 9\sin(y)\sin(x)\)

\(\displaystyle y' = \dfrac{9\sin(y)\sin(x)}{ 9\cos(x)\cos(y)}\) :?:

 

[Deleted member 4993]

Solving for \(\displaystyle y'\)

\(\displaystyle f(x) = 9\cos(x) \sin(y)= 7\) - Use product rule on left \(\displaystyle g(f') + f(g')\) once doing derivative.

\(\displaystyle f'(x) = [\sin(y)][-9\sin(x)] + [9\cos(x)][\cos(y)]y' = 0\)

\(\displaystyle f'(x) = -9\sin(y) \sin(x) + 9\cos(x)\cos(y)y' = 0\)

\(\displaystyle f'(x) = 9\cos(x)\cos(y)y' = -9\sin(y)\sin(x)\) .......................................Incorrect

\(\displaystyle y' = \dfrac{-9\sin(y)\sin(x)}{ 9\cos(x)\cos(y)}\) :?:

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[Jason76]

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Original Post Edited

 

[Deleted member 4993]

Post Edited

Solving for \(\displaystyle y'\)

\(\displaystyle f(x,y) = 9\cos(x) \sin(y)= 7\) - Use product rule on left \(\displaystyle g(f') + f(g')\) once doing derivative.

\(\displaystyle [\sin(y)][-9\sin(x)] + [9\cos(x)][\cos(y)]y' = 0\)

\(\displaystyle -9\sin(y) \sin(x) + 9\cos(x)\cos(y)y' = 0\)

\(\displaystyle 9\cos(x)\cos(y)y' = 9\sin(y)\sin(x)\)

\(\displaystyle y' = \dfrac{9\sin(y)\sin(x)}{ 9\cos(x)\cos(y)} \) ............ can be simplified further

\(\displaystyle y' = tan(x) * tan(y)\)
:?:

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