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Question Number 156105 by zainaltanjung last updated on 08/Oct/21

$$\mathrm{Find}\mathrm{The}\mathrm{derivatif}\mathrm{of}\mathrm{this}\mathrm{function}:$$$$1).{\mathrm{y}}^4+3\mathrm{y}-{4\mathrm{x}}^3=5\mathrm{x}+1$$$$2).{4\mathrm{xy}}^3-{\mathrm{x}}^2\mathrm{y}+{\mathrm{x}}^3-5\mathrm{x}+6=0$$$$3).{3\mathrm{y}}^4+4\mathrm{x}-{\mathrm{x}}^2\sin \mathrm{y}-4=0$$$$4).\mathrm{y}={\mathrm{x}}^2\sin \mathrm{y}$$$$5).{\sin }^{2}3\mathrm{y}=\mathrm{x}+\mathrm{y}-1$$$$$$

Answered by MathsFan last updated on 08/Oct/21

$$\{\begin{array}{l}1.{4\mathrm{y}}^3\frac{\mathrm{dy}}{\mathrm{dx}}+3\frac{\mathrm{dy}}{\mathrm{dx}}-{12\mathrm{x}}^2=5\to \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{5+{12\mathrm{x}}^2}{{4\mathrm{y}}^3+3}\\ 2.{12\mathrm{xy}}^2\frac{\mathrm{dy}}{\mathrm{dx}}-{\mathrm{x}}^2\frac{\mathrm{dy}}{\mathrm{dx}}=5+2\mathrm{yx}-{3\mathrm{x}}^2-{4\mathrm{y}}^3\to \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{5+2\mathrm{yx}-{3\mathrm{x}}^2-{4\mathrm{y}}^3}{{12\mathrm{xy}}^2-{\mathrm{x}}^2}\end{array}$$$$3.{12\mathrm{y}}^3\frac{\mathrm{dy}}{\mathrm{dx}}+4-{\mathrm{x}}^2\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{cosy}-2\mathrm{xsiny}=0$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\mathrm{xsiny}-4}{{12\mathrm{y}}^3-{\mathrm{x}}^2\mathrm{cosy}}$$$$$$$$$$

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