11. d d x e x + e x \frac{d}{dx} \sqrt{e^x}+e^{\sqrt{x}} d x d e x + e x
d d x e x + e x = ( e x 2 ) ′ + ( e x 1 2 ) ′ = e x 2 ( 1 2 ) + e x ( 1 2 x ) = 1 2 e x + e x 2 x
\begin{aligned}
&\frac{d}{dx} \sqrt{e^x}+e^{\sqrt{x}}\\
&=(e^{\frac{x}{2}})' + (e^{x^{\frac{1}{2}}})'\\
&=e^{\frac{x}{2}}(\frac{1}{2})+e^{\sqrt{x}}(\frac{1}{2\sqrt{x}})\\
&=\frac{1}{2}\sqrt{e^x}+\frac{e^{\sqrt{x}}}{2\sqrt{x}}
\end{aligned}
d x d e x + e x = ( e 2 x ) ′ + ( e x 2 1 ) ′ = e 2 x ( 2 1 ) + e x ( 2 x 1 ) = 2 1 e x + 2 x e x
12. d d x s e c 3 ( 2 x ) \frac{d}{dx} sec^3(2x) d x d s e c 3 ( 2 x )
d d x s e c 3 ( 2 x ) = 3 s e c 2 ( 2 x ) ( s e c ( 2 x ) t a n ( 2 x ) ) 2 = 6 s e c 3 ( 2 x ) t a n ( 2 x )
\begin{aligned}
&\frac{d}{dx} sec^3(2x)\\
&=3sec^2(2x)(sec(2x)tan(2x))2\\
&=6sec^3(2x)tan(2x)
\end{aligned}
d x d s e c 3 ( 2 x ) = 3 s e c 2 ( 2 x ) ( s e c ( 2 x ) t a n ( 2 x ) ) 2 = 6 s e c 3 ( 2 x ) t a n ( 2 x )
13. d d x 1 2 ( s e c x ) ( t a n x ) + 1 2 l n ( s e c x + t a n x ) \frac{d}{dx} \frac{1}{2} (secx)(tanx) + \frac{1}{2} ln(secx + tanx) d x d 2 1 ( s e c x ) ( t a n x ) + 2 1 l n ( s e c x + t a n x )
d d x 1 2 ( s e c x ) ( t a n x ) + 1 2 l n ( s e c x + t a n x ) = 1 2 ( s e c x t a n x t a n x + s e c x s e c 2 x ) + 1 2 ( s e c x + t a n x ) ( s e c x t a n x + s e c 2 x ) = 1 2 s e c x ( t a n 2 x + s e c 2 x ) + s e c x ( t a n x + s e c x ) 2 ( s e c x + t a n x ) = 1 2 s e c x ( 1 + t a n 2 x + s e c 2 x ) = 1 2 s e c x 2 s e c 2 x = s e c 3 x
\begin{aligned}
&\frac{d}{dx} \frac{1}{2} (secx)(tanx) + \frac{1}{2} ln(secx + tanx)\\
&=\frac{1}{2}(secxtanxtanx+secxsec^2x)+\frac{1}{2(secx+tanx)}(secxtanx+sec^2x)\\
&=\frac{1}{2}sec x(tan^2x+sec^2x)+\frac{secx(tanx+secx)}{2(secx+tanx)}\\
&=\frac{1}{2}secx(1+tan^2x+sec^2x)=\frac{1}{2}secx 2sec^2x\\
&=sec^3x
\end{aligned}
d x d 2 1 ( s e c x ) ( t a n x ) + 2 1 l n ( s e c x + t a n x ) = 2 1 ( s e c x t a n x t a n x + s e c x s e c 2 x ) + 2 ( s e c x + t a n x ) 1 ( s e c x t a n x + s e c 2 x ) = 2 1 s e c x ( t a n 2 x + s e c 2 x ) + 2 ( s e c x + t a n x ) s e c x ( t a n x + s e c x ) = 2 1 s e c x ( 1 + t a n 2 x + s e c 2 x ) = 2 1 s e c x 2 s e c 2 x = s e c 3 x
14. d d x ( x e x ) / ( 1 + e x ) \frac{d}{dx} (xe^x)/(1+e^x) d x d ( x e x ) / ( 1 + e x )
d d x x e x 1 + e x ( ( x e x ) ′ = e x + x e x ) = ( x e x ) ′ ( 1 + e x ) − ( x e x ) ( 1 + e x ) ′ ( 1 + e x ) 2 = ( e x + x e x ) ( 1 + e x ) − ( x e x ) ( e x ) ( 1 + e x ) 2 = e x ( 1 + x ) ( 1 + e x ) − x e 2 x ( 1 + e x ) 2 = e x ( ( 1 + x e x + e x + x ) − x e x ) ( 1 + e x ) 2 = e x ( 1 + e x + x ) ( 1 + e x ) 2
\begin{aligned}
&\frac{d}{dx} \frac{xe^x}{1+e^x}\\
&((xe^x)' = e^x+xe^x)\\
&=\frac{(xe^x)'(1+e^x) - (xe^x)(1+e^x)'}{(1+e^x)^2}\\
&=\frac{(e^x+xe^x)(1+e^x) - (xe^x)(e^x)}{(1+e^x)^2}\\
&=\frac{e^x(1+x)(1+e^x)-xe^{2x}}{(1+e^x)^2}\\
&=\frac{e^x((1+xe^x+e^x+x)-xe^{x})}{(1+e^x)^2}\\
&=\frac{e^x(1+e^x+x)}{(1+e^x)^2}\\
\end{aligned}
d x d 1 + e x x e x ( ( x e x ) ′ = e x + x e x ) = ( 1 + e x ) 2 ( x e x ) ′ ( 1 + e x ) − ( x e x ) ( 1 + e x ) ′ = ( 1 + e x ) 2 ( e x + x e x ) ( 1 + e x ) − ( x e x ) ( e x ) = ( 1 + e x ) 2 e x ( 1 + x ) ( 1 + e x ) − x e 2 x = ( 1 + e x ) 2 e x ( ( 1 + x e x + e x + x ) − x e x ) = ( 1 + e x ) 2 e x ( 1 + e x + x )
15. d d x ( e 4 x ) ( c o s ( x / 2 ) ) \frac{d}{dx} (e^{4x})(cos(x/2)) d x d ( e 4 x ) ( c o s ( x / 2 ) )
d d x ( e 4 x ) ( c o s ( x 2 ) ) = ( e 4 x ) ′ ( c o s ( x 2 ) ) + ( e 4 x ) ( c o s ( x 2 ) ) ′ = ( e 4 x 4 ) c o s ( x 2 ) + e 4 x ( − s i n ( x 2 ) 1 2 ) = 4 e 4 x c o s ( x 2 ) − 1 2 e 4 x s i n ( x 2 )
\begin{aligned}
&\frac{d}{dx} (e^{4x})(cos(\frac{x}{2}))\\
&=(e^{4x})'(cos(\frac{x}{2}))+(e^{4x})(cos(\frac{x}{2}))'\\
&=(e^{4x}4)cos(\frac{x}{2})+e^{4x}(-sin(\frac{x}{2})\frac{1}{2})\\
&=4e^{4x}cos(\frac{x}{2})-\frac{1}{2}e^{4x}sin(\frac{x}{2})
\end{aligned}
d x d ( e 4 x ) ( c o s ( 2 x ) ) = ( e 4 x ) ′ ( c o s ( 2 x ) ) + ( e 4 x ) ( c o s ( 2 x ) ) ′ = ( e 4 x 4 ) c o s ( 2 x ) + e 4 x ( − s i n ( 2 x ) 2 1 ) = 4 e 4 x c o s ( 2 x ) − 2 1 e 4 x s i n ( 2 x )
16. d d x 1 x 3 − 2 4 \frac{d}{dx} \frac{1}{\sqrt[4]{x^3 - 2}} d x d 4 x 3 − 2 1
d d x 1 x 3 − 2 4 = d d x ( x 3 − 2 ) − 1 4 = − 1 4 ( x 3 − 2 ) − 5 4 ( 3 x 2 ) = − 3 x 2 4 ( x 3 − 2 ) x 3 − 2 4 = − 3 x 2 4 ( x 3 − 2 ) 5 4
\begin{aligned}
&\frac{d}{dx} \frac{1}{\sqrt[4]{x^3 - 2}}=\frac{d}{dx} (x^3 - 2)^{-\frac{1}{4}}\\
&=-\frac{1}{4} (x^3 - 2)^{-\frac{5}{4}}(3x^2)\\
&=-\frac{3x^2}{4(x^3-2)\sqrt[4]{x^3 - 2}}\\
&=-\frac{3x^2}{4\sqrt[4]{(x^3 - 2)^5}}\\
\end{aligned}
d x d 4 x 3 − 2 1 = d x d ( x 3 − 2 ) − 4 1 = − 4 1 ( x 3 − 2 ) − 4 5 ( 3 x 2 ) = − 4 ( x 3 − 2 ) 4 x 3 − 2 3 x 2 = − 4 4 ( x 3 − 2 ) 5 3 x 2
17. d d x a r c t a n ( s q r t ( x 2 − 1 ) ) \frac{d}{dx} arctan(sqrt(x^2-1)) d x d a r c t a n ( s q r t ( x 2 − 1 ) )
d d x t a n − 1 ( x 2 − 1 ) ( y = a r c t a n ( x ) , t a n y = x , s e c 2 y d y = d x ) ( R . T a n g l e = y , a = 1 , o = x , h = 1 + x 2 ) ( d y / d x = 1 s e c 2 y = c o s 2 y = 1 1 + x 2 ) = c o s 2 ( t a n − 1 ( x 2 − 1 ) ) 1 2 x 2 − 1 2 x = c o s 2 ( t a n − 1 ( x 2 − 1 ) ) x x 2 − 1 = 1 1 + x 2 − 1 x x 2 − 1 = 1 x x 2 − 1
\begin{aligned}
&\frac{d}{dx} tan^{-1}(\sqrt{x^2-1})\\
& (y=arctan (x), tan y=x, sec^2ydy=dx)\\
&(R.T angle=y, a=1, o=x, h=\sqrt{1+x^2})\\
& (dy/dx = \frac{1}{sec^2y}=cos^2y=\frac{1}{1+x^2})\\
&=cos^2(tan^{-1}(\sqrt{x^2-1}))\frac{1}{2\sqrt{x^2-1}}2x\\
&=cos^2(tan^{-1}(\sqrt{x^2-1}))\frac{x}{\sqrt{x^2-1}}\\
&=\frac{1}{1+x^2-1}\frac{x}{\sqrt{x^2-1}}\\
&=\frac{1}{x\sqrt{x^2-1}}
\end{aligned}
d x d t a n − 1 ( x 2 − 1 ) ( y = a r c t a n ( x ) , t a n y = x , s e c 2 y d y = d x ) ( R . T a n g l e = y , a = 1 , o = x , h = 1 + x 2 ) ( d y / d x = s e c 2 y 1 = c o s 2 y = 1 + x 2 1 ) = c o s 2 ( t a n − 1 ( x 2 − 1 ) ) 2 x 2 − 1 1 2 x = c o s 2 ( t a n − 1 ( x 2 − 1 ) ) x 2 − 1 x = 1 + x 2 − 1 1 x 2 − 1 x = x x 2 − 1 1
18. d d x ( l n x ) / x 3 \frac{d}{dx} (lnx)/x^3 d x d ( l n x ) / x 3
d d x l n x x 3 = 1 x x 3 − l n x ( 3 x 2 ) x 6 = x 2 − 3 x 2 l n x x 6 = 1 − 3 l n x x 4
\begin{aligned}
&\frac{d}{dx} \frac{lnx}{x^3}=\frac{\frac{1}{x}x^3-lnx(3x^2)}{x^6}\\
&=\frac{x^2-3x^2lnx}{x^6}=\frac{1-3lnx}{x^4}
\end{aligned}
d x d x 3 l n x = x 6 x 1 x 3 − l n x ( 3 x 2 ) = x 6 x 2 − 3 x 2 l n x = x 4 1 − 3 l n x
19. d d x x x \frac{d}{dx} x^x d x d x x
d d x x x ( y = x x , l o g x y = x , l n y l n x = x ) ( l n y = x l n x , 1 y d y = ( l n x + 1 ) d x ) d y d x = ( 1 + l n x ) y = ( 1 + l n x ) x x
\begin{aligned}
&\frac{d}{dx} x^x\\
&(y=x^x, log_xy=x, \frac{ln y}{ln x}=x)\\
&(lny = xlnx, \frac{1}{y}dy=(lnx+1)dx)\\
&\frac{dy}{dx}=(1+lnx)y=(1+lnx)x^x
\end{aligned}
d x d x x ( y = x x , l o g x y = x , l n x l n y = x ) ( l n y = x l n x , y 1 d y = ( l n x + 1 ) d x ) d x d y = ( 1 + l n x ) y = ( 1 + l n x ) x x d d x x x , ( x = e l n x ) = d d x ( e l n x ) x = d d x ( e x l n x ) = e x l n x ( l n x + 1 ) = x x ( l n x + 1 )
\frac{d}{dx} x^x, (x=e^{lnx})\\
=\frac{d}{dx} (e^{lnx})^x=\frac{d}{dx} (e^{xlnx})\\
=e^{xlnx}(lnx+1)\\
=x^x(lnx+1)
d x d x x , ( x = e l n x ) = d x d ( e l n x ) x = d x d ( e x l n x ) = e x l n x ( l n x + 1 ) = x x ( l n x + 1 )
20. d d x ( x 3 + y 3 = 6 x y ) \frac{d}{dx}(x^3+y^3=6xy) d x d ( x 3 + y 3 = 6 x y )
d d x ( x 3 + y 3 = 6 x y ) 3 x 2 d x + 3 y 2 d y = 6 y d x + 6 x d y ( 3 x 2 − 6 y ) d x = ( 6 x − 3 y 2 ) d y d y d x = 3 x 2 − 6 y 6 x − 3 y 2 = x 2 − 2 y 2 x − y 2
\begin{aligned}
&\frac{d}{dx}(x^3+y^3=6xy) \\
&3x^2dx+3y^2dy=6ydx+6xdy\\
&(3x^2-6y)dx=(6x-3y^2)dy\\
&\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}\\
&=\frac{x^2-2y}{2x-y^2}
\end{aligned}
d x d ( x 3 + y 3 = 6 x y ) 3 x 2 d x + 3 y 2 d y = 6 y d x + 6 x d y ( 3 x 2 − 6 y ) d x = ( 6 x − 3 y 2 ) d y d x d y = 6 x − 3 y 2 3 x 2 − 6 y = 2 x − y 2 x 2 − 2 y x 3 + y 3 = 6 x y : ( D x ) 3 x 2 + 3 y 2 y ′ = 6 y + 6 x y ′ ( 3 x 2 − 6 y ) = ( 6 x − 3 y 2 ) y ′ y ′ = 3 x 2 − 6 y 6 x − 3 y 2 = x 2 − 2 y 2 x − y 2
\begin{aligned}
&x^3+y^3=6xy :(Dx)\\
&3x^2+3y^2y'=6y+6xy'\\
&(3x^2-6y)=(6x-3y^2)y'\\
&y'=\frac{3x^2-6y}{6x-3y^2}\\
&=\frac{x^2-2y}{2x-y^2}
\end{aligned}
x 3 + y 3 = 6 x y : ( D x ) 3 x 2 + 3 y 2 y ′ = 6 y + 6 x y ′ ( 3 x 2 − 6 y ) = ( 6 x − 3 y 2 ) y ′ y ′ = 6 x − 3 y 2 3 x 2 − 6 y = 2 x − y 2 x 2 − 2 y
Author: crazyj7@gmail.com
