Question
Asked Oct 23, 2019
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Find an equation of the line tangent to the curve defined by

x^3+6xy+y^5=21 at the point (2,1)

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Expert Answer

Step 1

Step 2

Differentiate the given equation with respect to x.

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d - 6ay + y' = 21) dx 3x2+6x6+5y* 0=

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Step 3

Now substitute the given point (2, 1) in the above obtained ...

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3(2)+6(2)6(1)+5(1)* . (12+5)12+6 0 17 18 dy 18 17

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Calculus

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