39) y = (x^2+2)^2 (x^4+4)^4
ln y = ln[(x^2+2)^2(x^4+4)^4]
ln y = ln(x^2+2)^2 + ln(x^4+4)^4
ln y

Name: 39) y = (x^2+2)^2 (x^4+4)^4 ln y = ln[(x^2+2)^2(x^4+4)^4] ln y = ln(x^2+2)^2 + ln(x^4+4)^4 ln y = 2ln(x^2+2) + 4ln(x^4+4) fracy'y = 2frac1(x^2+2) cdot 2x + 4frac1(x^4+4) cdot 4x^3 y' = y(frac4xx^2+2 + frac16x^3x^4+4)
Uploaded: 2023-02-09T12:02:40-08:00
Duration: 1 min 19 s
Channel: Madhur L
Description: 39) y = (x^2+2)^2 (x^4+4)^4 ln y = ln[(x^2+2)^2(x^4+4)^4] ln y = ln(x^2+2)^2 + ln(x^4+4)^4 ln y = 2ln(x^2+2) + 4ln(x^4+4) fracy'y = 2frac1(x^2+2) cdot 2x + 4frac1(x^4+4) cdot 4x^3 y' = y(frac4xx^2+2 + frac16x^3x^4+4)

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Madhur L · Accepted Answer

Hi, y is given to us as x squared plus two whole square x to the power 4 plus 4 whole to the pow

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39) $y = (x^2+2)^2 (x^4+4)^4$ $ln y = ln[(x^2+2)^2(x^4+4)^4]$ $ln y = ln(x^2+2)^2 + ln(x^4+4)^4$ $ln y = 2ln(x^2+2) + 4ln(x^4+4)$ $frac{y'}{y} = 2frac{1}{(x^2+2)} cdot 2x + 4frac{1}{(x^4+4)} cdot 4x^3$ $y' = y(frac{4x}{x^2+2} + frac{16x^3}{x^4+4})$

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