How do you implicitly differentiate $ysinx^2 = xsin(e^y)$ ?

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Answer 1 · 2018-04-02T12:45:11

$dy/dx$ = $(sin(e^y)-2xycos(x^2))/(sin(x^2)-e^yxcos(e^y))$

$y times 2xcos(x^2) + sin(x^2) times dy/dx$ = $x times e^ycos(e^y) times dy/dx + sin(e^y)$

$2xycos(x^2) + dy/dx times sin(x^2)$ = $dy/dx times e^yxcos(e^y) + sin (e^y)$

$dy/dx sin (x^2) - e^ycos(e^y)$ = $sin (e^y) - 2xycos(x^2)$

$dy/dx$ = $(sin(e^y)-2xycos(x^2))/(sin(x^2)-e^yxcos(e^y))$

How do you implicitly differentiate ysinx^2 = xsin(e^y) ysinx^2 = xsin(e^y)?