Apply the product rule: 

\(\displaystyle \frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]\) 

Then apply the product rule \(\displaystyle nx^{n-1}\) (where "n" is the exponent) where needed.

\(\displaystyle \\ \frac{d}{dx}(\sqrt{x}sin(x))\\ \\=\frac{d}{dx}(x^{\frac{1}{2}}sin(x))\\ \\=x^{\frac{1}{2}}\frac{d}{dx}(sin(x))+sin(x)\frac{d}{dx}(x^{\frac{1}{2}})\\ \\=x^{\frac{1}{2}}cos(x)+sin(x)\frac{1}{2}x^{\frac{1}{2}-1}\\ \\=x^{\frac{1}{2}}cos(x)+\frac{1}{2}sin(x)x^{-\frac{1}{2}}\)

Apply the product rule: 

\(\displaystyle \frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]\) 

Then apply the product rule \(\displaystyle nx^{n-1}\) (where "n" is the exponent) where needed.

\(\displaystyle \\ \frac{d}{dx}\left(\left(\frac{1}{x^2}-\frac{3}{x^4}\right)(x+5x^3)\right)\\ \\=(x^{-2}-3x^{-4})\frac{d}{dx}(x+5x^3)+(x+5x^3)\frac{d}{dx}(x^{-2}-3x^{-4})\\ \\=(x^{-2}-3x^{-4})(1*x^{1-1}+3*5x^{3-1})+(x+5x^3)(-2x^{-2-1}-(-4)*3x^{-4-1})\\ \\=(x^{-2}-3x^{-4})(1+15x^2)+(x+5x^3)(-2x^{-3}+12x^{-5})\)

Apply the product rule: 

\(\displaystyle \frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]\) 

Then apply the product rule \(\displaystyle nx^{n-1}\) (where "n" is the exponent) where needed.

\(\displaystyle \\ \frac{d}{dx}[(x^3-2x)(x^{-4}+x^{-2})]\\ \\=(x^3-2x)\frac{d}{dx}(x^{-4}+x^{-2})+(x^{-4}+x^{-2})\frac{d}{dx}(x^3-2x)\\ \\=(x^3-2x)((-4)x^{-4-1}+(-2)x^{-2-1})+(x^{-4}+x^{-2})(3x^{3-1}-2x^{1-1})\\ \\=(x^3-2x)(-4x^{-5}-2x^{-3})+(x^{-4}+x^{-2})(3x^2-2)\)

Because this is a sum of functions, take the derivate of each function with respect to "x" and add them together:

\(\displaystyle \\ \frac{d}{dx}[sin(x)+cot(x)]\\ \\=\frac{d}{dx}(sin(x))+\frac{d}{dx}(cot(x))\\ \\=cos(x)+(-csc^(x))\\ \\=cos(x)-csc^2(x)\)

The derivative of \(\displaystyle sin(x)\) is \(\displaystyle cos(x)\) and the derivate of \(\displaystyle cot(x)\) is \(\displaystyle -csc^2(x)\).

Apply the product rule: 

\(\displaystyle \frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]\) 

Then apply the product rule \(\displaystyle nx^{n-1}\) (where "n" is the exponent) where needed.

\(\displaystyle \\ \frac{d}{dx}[x^2sin(x)]\\ \\=x^2\frac{d}{dx}(sin(x))+sin(x)\frac{d}{dx}(x^2)\\ \\=x^2cos(x)+sin(x)2x^{2-1}\\ \\=x^2cos(x)+2xsin(x)\)

Apply the product rule: 

\(\displaystyle \frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]\) 

Then apply the product rule \(\displaystyle nx^{n-1}\) (where "n" is the exponent) where needed.

\(\displaystyle \\ \frac{d}{dx}[(x^2+2)(x^{\frac{1}{2}}-1)]\\ \\=(x^2+2)\frac{d}{dx}(x^{\frac{1}{2}}-1)+(x^{\frac{1}{2}}-1)\frac{d}{dx}(x^2+2)\\ \\=(x^2+2)(\frac{1}{2}x^{\frac{1}{2}-1}-0)+(x^{\frac{1}{2}}-1)(2x^{2-1}+0)\\ \\=\frac{1}{2}x^{-\frac{1}{2}}(x^2+2)+2x(x^{\frac{1}{2}}-1)\\ \\=\frac{1}{2}x^{\frac{3}{2}}+x^{-\frac{1}{2}}+2x^{\frac{3}{2}}-2x\\ \\=\frac{5}{2}x^{\frac{3}{2}}-2x+x^{-\frac{1}{2}}\)

Apply the product rule: 

\(\displaystyle \frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]\)

Let \(\displaystyle f(x)=sin(x)\) and \(\displaystyle g(x)=cos(x)\) 

\(\displaystyle \\ \frac{d}{dx}[\sin (x)\cos (x)]\\ \\=\sin (x)\frac{d}{dx}\cos (x)+\cos (x)\frac{d}{dx}\sin (x)\\ \\=-\sin (x)\sin (x)+\cos (x)\cos (x)\\ \\=\cos^2(x)-\sin^2(x)\)

Apply the product rule: 

\(\displaystyle \frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]\) 

Then apply the product rule \(\displaystyle nx^{n-1}\) (where "n" is the exponent) where needed.

\(\displaystyle \\ \frac{d}{dx}[(x^3+5x)(x^{-2}-6x^2)]\\ \\=(x^3+5x)\frac{d}{dx}(x^{-2}-6x^2)+(x^{-2}-6x^2)\frac{d}{dx}(x^3+5x)\\ \\=(x^3+5x)(-2x^{-2-1}-(2)6x^{2-1})+(x^{-2}-6x^2)(3x^{3-1}+5x^{1-1})\\ \\=(x^3+5x)(-2x^{-3}-12x)+(x^{-2}-6x^2)(3x^2+5)\)

Apply the product rule: 

\(\displaystyle \frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]\) 

Then apply the product rule \(\displaystyle nx^{n-1}\) (where "n" is the exponent) where needed.

\(\displaystyle \\\frac{d}{dx}[(x^2+x)\cos (x)]\\ \\=(x^2+x)\frac{d}{dx}\cos (x)+\cos (x)\frac{d}{dx}(x^2+x)\\ \\=(x^2+x)(-\sin (x))+\cos (x)(2x^{2-1}+1x^{^{1-1}})\\ \\=-(x^2+x)\sin (x)+(2x+1)\cos (x)\)

Apply the product rule: 

\(\displaystyle \frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]\)  

\(\displaystyle \\ \frac{d}{dx}[(\cos (x)+\sin (x))(\sin (x)-\cos (x))]\\ \\=(\cos (x)+\sin (x))\frac{d}{dx}(\sin (x)-\cos (x))+(\sin (x)-\cos (x))\frac{d}{dx}(\cos (x)+\sin (x))\\ \\=(\cos (x)+\sin (x))(\cos (x)-(-\sin (x)))+(\sin (x)-\cos (x))(-\sin (x)+\cos (x))\\ \\=(\cos (x)+\sin (x))^2+(\sin (x)-\cos (x))(\cos (x)-\sin (x))\)