Find ππ¦
ππ₯
in each case
(a) π¦ = π‘ππ4(3π₯)
(3)
(b) π₯2π¦2 + π₯π πππ¦ = 4 use implicit differentiation
Let us find "\\frac{dy}{dx}" in each case
(a) "\ud835\udc66 = \\tan^4(3\ud835\udc65)"
(b) "\ud835\udc65^2\ud835\udc66^2 + \ud835\udc65\\sin\ud835\udc66 = 4" use implicit differentiation
(a) "\ud835\udc66' = 4\\tan^3(3\ud835\udc65)(\\tan(3\ud835\udc65))'= 4\\tan^3(3\ud835\udc65)\\frac{1}{\\cos^2 (3x)}(3\ud835\udc65)'=\n12\\frac{\\tan^3(3\ud835\udc65)}{\\cos^2 (3x)}."
(b) Let us use the implicit differentiation:
"2x\ud835\udc66^2 + \ud835\udc65^22\ud835\udc66y' +\\sin\ud835\udc66+xy'\\cos y = 0."
Therefore, "y'(2\ud835\udc65^2y+x\\cos y) = -2x\ud835\udc66^2 - \\sin y," and hence
"y' =- \\frac{2x\ud835\udc66^2 +\\sin y}{2\ud835\udc65^2y+x\\cos y}."
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