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Asked Nov 8, 2019
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Find the equation of the tangent line at the given point on the curve.
109y2e 327x/10. (10,3)
, 3x2 + Зу2
= xe
e
у3
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Find the equation of the tangent line at the given point on the curve. 109y2e 327x/10. (10,3) , 3x2 + Зу2 = xe e у3

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Expert Answer

Step 1

Given,

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327x/10 109y ye = xe 109y32+3y2 хе 327x/10 e 109y 3x+3y хе -e 327x/10 e

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Step 2

Differentiate the function y with respect to x

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3x2+3y2 109y хе -e 327x/10 dx e df(u)_ df_du Apply the chain rule : dx du d 109y 3x2+3y2 хе 1 - e f=u2. 327x 10 109y 3r2+3y2 хе и2 du 327x 10 327x+30x2+30y2 327x+1090 y 327x+3x2+32 -60e 10 327x+109y 10e 10 10 10 x-327e 327x 1 x+327e 5 2u2 10e

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Step 3

Substitute b...

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327x+30x2+30y2 327x+3x2+32 327x+109 y 327x+1090y 10 10 dy10e 10 60e 10 x-327e x+327e dx 981x 3x2+31,2 2 109 y. X-e 20e 20 327x43x2+3y2 327x+30x2+30y2 10 327x109y 327x+1090y 10 x-327e x+327e 10e 10 -60e 10 981x 3r2+3y2 20e 2009yx-e

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Math

Calculus