Derivatives

[paulxzt]

Can someone please check my work and see if it's correct? Thank you

find deriv. of y = 2sqrt(x) - (1)/2sqrt(x)

y' = 2x^1/2 - 1/2(x)^1/2

x^-1/2 - ( 2x^1/2)(0) - 1(x^-1/2)
-----------------------------------------
4x

= 1 / sqrt(x) + 1/ 4xsqrt(x)

Sorry if the format is hard to read. Any help/suggestion is appreciated

2) y = sin^2(3x) + cos^2(3x)

can i rewrite this as (sin3x)^2 + (cos3x)^2?

would the derivative come out to 0 ?

 

[soroban]

Hello, Paul!

You're doing great!

Find deriv. of: \(\displaystyle \L\,y \:= \:2\sqrt{x}\,-\,\frac{1}{2\sqrt{x}}\)

\(\displaystyle \L y \:= \:2x^{\frac{1}{2}}\,-\,\frac{1}{2}x^{-\frac{1}{2}}\)

\(\displaystyle \L y' \;=\;2\cdot\frac{1}{2}x^{-\frac{1}{2}} \,-\,\frac{1}{2}\left(-\frac{1}{2}\right)x^{-\frac{3}{2}} \;= \;x^{-\frac{1}{2}}\,+\,\frac{1}{4}x^{-\frac{3}{2}} \;= \;\frac{1}{\sqrt{x}} \,+\,\frac{1}{4x\sqrt{x}}\) . . . You're right!

\(\displaystyle 2)\;\;y \:= \:\sin^2(3x)\,+\,\cos^2(3x)\)

Can i rewrite this as: \(\displaystyle \, (\sin3x)^2 + (\cos3x)^2\) ? . . . Yes!

Would the derivative come out to 0 ? . . . Correct!

They were having their little joke . . .

Did you notice that: \(\displaystyle \,\sin^2(3x)\,+\,\cos^2(3x) \:=\:1\) ?