1. Rewrite the expression in terms of logarithms:
[tex]y=x^x=e^{\ln x^x}=e^{x\ln x}[/tex]
Then differentiate with the chain rule (I'll use prime notation to save space; that is, the derivative of y is denoted y' )
[tex]y'=e^{x\ln x}(x\ln x)'=x^x(x\ln x)'[/tex]
[tex]y'=x^x(x'\ln x+x(\ln x)')[/tex]
[tex]y'=x^x\left(\ln x+\dfrac xx\right)[/tex]
[tex]y'=x^x(\ln x+1)[/tex]
2. Chain rule:
[tex]y=\ln(\csc(3x))[/tex]
[tex]y'=\dfrac1{\csc(3x)}(\csc(3x))'[/tex]
[tex]y'=\sin(3x)\left(-\cot^2(3x)(3x)'\right)[/tex]
[tex]y'=-3\sin(3x)\cot^2(3x)[/tex]
Since [tex]\cot x=\frac{\cos x}{\sin x}[/tex], we can cancel one factor of sine:
[tex]y'=-3\dfrac{\cos^2(3x)}{\sin(3x)}=-3\cos(3x)\cot(3x)[/tex]
3. Chain rule:
[tex]y=e^{e^{\sin x}}[/tex]
[tex]y'=e^{e^{\sin x}}\left(e^{\sin x}\right)'[/tex]
[tex]y'=e^{e^{\sin x}}e^{\sin x}(\sin x)'[/tex]
[tex]y'=e^{e^{\sin x}+\sin x}\cos x[/tex]
4. If you're like me and don't remember the rule for differentiating logarithms of bases not equal to e, you can use the change-of-base formula first:
[tex]\log_2x=\dfrac{\ln x}{\ln2}[/tex]
Then
[tex](\log_2x)'=\left(\dfrac{\ln x}{\ln 2}\right)'=\dfrac1{\ln 2}[/tex]
So we have
[tex]y=\cos^2(\log_2x)[/tex]
[tex]y'=2\cos(\log_2x)\left(\cos(\log_2x)\right)'[/tex]
[tex]y'=2\cos(\log_2x)(-\sin(\log_2x))(\log_2x)'[/tex]
[tex]y'=-\dfrac2{\ln2}\cos(\log_2x)\sin(\log_2x)[/tex]
and we can use the double angle identity and logarithm properties to condense this result:
[tex]y'=-\dfrac1{\ln2}\sin(2\log_2x)=-\dfrac1{\ln2}\sin(\log_2x^2)[/tex]
5. Differentiate both sides:
[tex]\left(x^2-y^2+\sin x\,e^y+\ln y\,x\right)'=0'[/tex]
[tex]2x-2yy'+\cos x\,e^y+\sin x\,e^yy'+\dfrac{xy'}y+\ln y=0[/tex]
[tex]-\left(2y-\sin x\,e^y-\dfrac xy\right)y'=-\left(2x+\cos x\,e^y+\ln y\right)[/tex]
[tex]y'=\dfrac{2x+\cos x\,e^y\ln y}{2y-\sin x\,e^y-\frac xy}[/tex]
[tex]y'=\dfrac{2xy+\cos x\,ye^y\ln y}{2y^2-\sin x\,ye^y-x}[/tex]
6. Same as with (5):
[tex]\left(\sin(x^2+\tan y)+e^{x^3\sec y}+2x-y+2\right)'=0'[/tex]
[tex]\cos(x^2+\tan y)(x^2+\tan y)'+e^{x^3\sec y}(x^3\sec y)'+2-y'=0[/tex]
[tex]\cos(x^2+\tan y)(2x+\sec^2y y')+e^{x^3\sec y}(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0[/tex]
[tex]\cos(x^2+\tan y)(2x+\sec^2y y')+e^{x^3\sec y}(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0[/tex]
[tex]\left(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1\right)y'=-\left(2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2\right)[/tex]
[tex]y'=-\dfrac{2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2}{\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1}[/tex]
7. Looks like
[tex]y=x^2-e^{2x}[/tex]
Compute the second derivative:
[tex]y'=2x-2e^{2x}[/tex]
[tex]y''=2-4e^{2x}[/tex]
Set this equal to 0 and solve for x :
[tex]2-4e^{2x}=0[/tex]
[tex]4e^{2x}=2[/tex]
[tex]e^{2x}=\dfrac12[/tex]
[tex]2x=\ln\dfrac12=-\ln2[/tex]
[tex]x=-\dfrac{\ln2}2[/tex]
