1. Find the derivatives of the following functions with respect to x.
π₯3 + π¦3 = 3
π¦ = (sin π₯)π‘πππ₯
π₯3+π¦3=3π₯^3 + π¦^3 = 3x3+y3=3
y=(3βx3)1/3y=(3-x^3)^{1/3}y=(3βx3)1/3
3x2+3y2yβ²=03x^2+3y^2y'=03x2+3y2yβ²=0
yβ²=βx2y2=βx2(3βx3)2/3y'=-\frac{x^2}{y^2}=-\frac{x^2}{(3-x^3)^{2/3}}yβ²=βy2x2β=β(3βx3)2/3x2β
π¦=(sinπ₯)π‘πππ₯π¦ = (sin π₯)^{π‘πππ₯}y=(sinx)tanx
lny=tanxln(sinx)lny=tanxln(sinx)lny=tanxln(sinx)
(lny)β²=yβ²y=ln(sinx)cos2x+tanxcosxsinx=ln(sinx)cos2x+1(lny)'=\frac{y'}{y}=\frac{ln(sinx)}{cos^2x}+\frac{tanxcosx}{sinx}=\frac{ln(sinx)}{cos^2x}+1(lny)β²=yyβ²β=cos2xln(sinx)β+sinxtanxcosxβ=cos2xln(sinx)β+1
yβ²=(ln(sinx)cos2x+1)y=(ln(sinx)cos2x+1)(sinπ₯)π‘πππ₯y'=(\frac{ln(sinx)}{cos^2x}+1)y=(\frac{ln(sinx)}{cos^2x}+1)(sin π₯)^{π‘πππ₯}yβ²=(cos2xln(sinx)β+1)y=(cos2xln(sinx)β+1)(sinx)tanx
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