Answer to Question #315308 in Calculus for Jimjim

Question #315308

Differentiate the following functions:



1.y=1/2(3x^2+1)^2



2.y=3+4x-x^2



Find the derivative of the following



1.y=√x^2-2/2x^3+1



2.y=x^2√6x^2/3x



3.y=2x^2√x

1
Expert's answer
2022-03-22T18:35:24-0400

1)y=12(3x2+1)2y=12(2)1(3x2+1)3(6x)==6x(3x2+1)22)y=3+4xx2y=0+42x=42x3)y=x222x3+1y=12x222x(2x3+1)x226x2(2x3+1)2==x(2x3+1)(x22)6x2x22(2x3+1)2==4x4+12x2+xx22(2x3+1)24)y=x26x23xy=x6x23=63x2y=632x=26x35)y=2x2x=2x52y=252x32=5x321) y=\frac{1}{2(3x^2+1)^2}\\ y'=\frac{1}{2}(-2)\frac{1}{(3x^2+1)^3}\cdot(6x)=\\ =\frac{-6x}{(3x^2+1)^2}\\ 2) y=3+4x-x^2\\ y'=0+4-2x=4-2x\\ 3) y=\frac{\sqrt{x^2-2}}{2x^3+1}\\ y'=\frac{\frac{1}{2\sqrt{x^2-2}}\cdot 2x\cdot (2x^3+1)- \sqrt{x^2-2}\cdot 6x^2}{(2x^3+1)^2}=\\ =\frac{x\cdot (2x^3+1)- (x^2-2)\cdot 6x^2}{\sqrt{x^2-2}(2x^3+1)^2}=\\ =\frac{-4x^4+12x^2+x}{\sqrt{x^2-2}(2x^3+1)^2}\\ 4) y=\frac{x^2\sqrt{6x^2}}{3x}\\ y=\frac{x\sqrt{6x^2}}{3}=\frac{\sqrt{6}}{3}x^2\\ y'=\frac{\sqrt{6}}{3}2x=\frac{2\sqrt{6}x}{3}\\ 5) y=2x^2\sqrt{x}=2x^{\frac{5}{2}}\\ y'=2\cdot\frac{5}{2}\cdot x^{\frac{3}{2}}=5x^{\frac{3}{2}}


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