Why am i so lost can someone pls help | Mathematics | Page 1

sullen mural Dec 15, 2022, 10:13 PM

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I keep getting the wrong answer

tough pier Dec 15, 2022, 10:46 PM

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You should use logarithmic differentiation

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https://www.youtube.com/watch?v=9z1Dz60mWcQ

YouTube

The Organic Chemistry Tutor

Introduction to Logarithmic Differentiation

This calculus video tutorial provides a basic introduction into logarithmic differentiation. It explains how to find the derivative of functions such as x^x, x^sinx, (lnx)^x, and x^(1/x). You need to take the natural log of both sides of the equation and perform implicit differentiation to find dy/dx. You need to be familar with product rule ...

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crude gate Dec 16, 2022, 3:53 AM

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(1) y=x^(x^3)
y'=x^(x^3) * (3(x^2) * ln(x)+(x^3)*(1/x)), generalized power rule + product rule
y'=x^((x^3)+2) * (3ln(x)+1)

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(2) d/dx[y=cos^x(x)]
y'=cos^x(x)*d/dx[ln(cos(x))x], product rule&chain
y'=cos^x(x)*( 1*ln(cos(x))+x*1/cos(x) * d/dx[cos(x)]), cos derivitive=-sin
y'=cos^x(x) * (ln(cos(x))-(xsin(x)/cos(x)))

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y'=cos^x(x)(ln(cos(x))-xtan(x))

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(3) d/dx[4-2y^2=4x^3]

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d/dx[4-2y^2]=d/dx[4x^3]

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≡d/dx[4]-2*d/dx[y^2]=4*d/dx[x^3], as derivitive is linear

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≡0-4(dy/dx)=12x^2

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≡-4yy'=12x^2

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=> y' = -3(x^2)/y

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now for the second derivative

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d/dx[y']=d/dx[-3(x^2)/y]

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≡y''=-3*d/dx[x^2/y]

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≡y''=-(3(2xy-y'x^2)/y^2

#Why am i so lost can someone pls help