Answer to Question #108255 in Calculus for please

Question

[img]https://upload.cc/i1/2020/04/06/jDxf0z.jpg[/img]



the question is in the photo

Answer 1

(a)$\frac{dy}{dx}=y'$

(i)

$y=\ln[\cos^{-1}((R+2)x)]\\ y'=\frac{1}{\cos^{-1}((R+2)x)}\cdot(-1)\cos^{-2}((R+2)x)\cdot\\ \cdot(-\sin((R+2)x))\cdot(R+2)=\\ =\tan((R+2)x)\cdot (R+2)\\ \frac{dy}{dx}=\tan((R+2)x)\cdot (R+2)$

(ii)

$x^3-2y^2=\tan(x^2y^3)\\ 3x^2-2\cdot2y\cdot y'=\frac{1}{\cos^2(x^2y^3)}\cdot(2xy^3+x^2\cdot3y^2\cdot y')\\ 3x^2-4y\cdot y'=\frac{1}{\cos^2(x^2y^3)}\cdot(2xy^3+3x^2y^2\cdot y')\\ (3x^2-4y\cdot y')\cos^2(x^2y^3)=2xy^3+3x^2y^2\cdot y'\\ y'=\frac{2xy^3-3x^2\cdot\cos^2(x^2y^3)}{-4y\cos^{2}(x^2y^3)-3x^2y^2}\\ \frac{dy}{dx}=\frac{2xy^3-3x^2\cdot\cos^2(x^2y^3)}{-4y\cos^{2}(x^2y^3)-3x^2y^2}$

(iii)

$y=(1+x^2)^{\sqrt{x}}\\ \ln y=\ln((1+x^2)^{\sqrt{x}})\\ \ln y=\sqrt{x}\ln(1+x^2)\\ \frac{1}{y}\cdot y'=\frac{1}{2\sqrt{x}}ln(1+x^2)+\sqrt{x}\cdot\frac{1}{1+x^2}\cdot 2x\\ y'=(\frac{1}{2\sqrt{x}}ln(1+x^2)+\sqrt{x}\cdot\frac{2x}{1+x^2})(1+x^2)^{\sqrt{x}}\\ \frac{dy}{dx}=(\frac{1}{2\sqrt{x}}ln(1+x^2)+\sqrt{x}\cdot\frac{2x}{1+x^2})(1+x^2)^{\sqrt{x}}$

(b)

$f(x)=x^5+4x+8\\ f'(x)=5x^4+4\\ x^5+4x+8=3\\ x^5+4x+5=0\\ x=-1\\ (-1)^5+4(-1)+5=-1-4+5=0\\ (f^{-1})'(3)=\frac{1}{f'(-1)}\\ f'(-1)=5\cdot(-1)^4+4=5+4=9\\ (f^{-1})'(3)=\frac{1}{9}$

(c)

$g(x)=(1-x)(1-2x)...(1-nx)\\ g'(x)=-(1-2x)(1-3x)...(1-nx)+\\ +(-2)(1-x)(1-3x)...(1-nx)+\\ +...+(-n)(1-x)(1-2x)...(1-(n-1)x)\\ g'(0)=-1+(-2)+...+(-n)=\\ =-(1+2+...+n)=-\frac{1+n}{2}\cdot n=-\frac{n^2+n}{2}$