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Question Number 91476 by Zainal Arifin last updated on 01/May/20

$$\mathrm{Find}\mathrm{the}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{tangent}$$$$\mathrm{line}\mathrm{to}\mathrm{the}\mathrm{graph}\mathrm{of}:$$$${\mathrm{y}}^4+3\mathrm{y}-{4\mathrm{x}}^3=5\mathrm{x}+1\mathrm{at}\mathrm{the}\mathrm{point}$$$$\mathrm{P}(1,-2)$$

Commented by john santu last updated on 01/May/20

$$(4y^3+3)y^{\prime }=12x^2+5$$$$y^{\prime }=\frac{12x^2+5}{4y^3+3}\Rightarrow slope=-\frac{17}{29}$$

Commented by Zainal Arifin last updated on 01/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir},\mathrm{you}\mathrm{are}\mathrm{right}.$$

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