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ddx[f(x)+g(x)]=ddxf(x)+ddxg(x)\frac{d}{dx}\left[f\left(x\right)+g\left(x\right)\right]=\frac{d}{dx}f\left(x\right)+\frac{d}{dx}g\left(x\right)dxd[f(x)+g(x)]=dxdf(x)+dxdg(x) is called
Product rule
sum rule
reciprocal rule
ddx[cg(x)]=\frac{d}{dx}\left[cg\left(x\right)\right]=dxd[cg(x)]=
c+ddxg(x)c+\frac{d}{dx}g\left(x\right)c+dxdg(x)
ddxf(x)\frac{d}{dx}f\left(x\right)dxdf(x)
c ddxg(x)c\ \frac{d}{dx}g\left(x\right)c dxdg(x)
Is the sentence true or false ?
(ddx[f(x)g(x)]=ddx[f(x)]g(x)−f(x) ddx[g(x)])[g(x)]2\frac{\left(\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{d}{dx}\left[f\left(x\right)\right]g\left(x\right)-f\left(x\right)\ \frac{d}{dx}\left[g\left(x\right)\right]\right)}{\left[g\left(x\right)\right]^2}[g(x)]2(dxd[g(x)f(x)]=dxd[f(x)]g(x)−f(x) dxd[g(x)]) is called sum rule
True
False
Given c is constant,dcdx=\frac{dc}{dx}=dxdc=
c+1
0
c
ddx[1g(x)]=\frac{d}{dx}\left[\frac{1}{g\left(x\right)}\right]=dxd[g(x)1]=
−ddx[g(x)][g(x)]2-\frac{\frac{d}{dx}\left[g\left(x\right)\right]}{\left[g\left(x\right)\right]^2}−[g(x)]2dxd[g(x)]
ddx[g(x)][g(x)]2=\frac{\frac{d}{dx}\left[g\left(x\right)\right]}{\left[g\left(x\right)\right]^2}=[g(x)]2dxd[g(x)]=
ddxf(x)+ddxg(x)=\frac{d}{dx}f\left(x\right)+\frac{d}{dx}g\left(x\right)=dxdf(x)+dxdg(x)=
ddx[x]=\frac{d}{dx}\left[x\right]=dxd[x]=
1
x
Given k is constant ddy[(ym)]=\frac{d}{dy}\left[\left(y^m\right)\right]=dyd[(ym)]=
n(ym+1)n\left(y^{m+1}\right)n(ym+1)
n(yn)n\left(y^n\right)n(yn)
m(ym−1)m\left(y^{m-1}\right)m(ym−1)
ddx[1g(x)]=−ddx[g(x)][g(x)]2\frac{d}{dx}\left[\frac{1}{g\left(x\right)}\right]=-\frac{\frac{d}{dx}\left[g\left(x\right)\right]}{\left[g\left(x\right)\right]^2}dxd[g(x)1]=−[g(x)]2dxd[g(x)] is called
ddx[h(x)−g(x)]=\frac{d}{dx}\left[h\left(x\right)-g\left(x\right)\right]=dxd[h(x)−g(x)]=
ddxh(x)−ddxp(x)\frac{d}{dx}h\left(x\right)-\frac{d}{dx}p\left(x\right)dxdh(x)−dxdp(x)
ddxh(x)\frac{d}{dx}h\left(x\right)dxdh(x)
ddxh(x)+ddxp(x)\frac{d}{dx}h\left(x\right)+\frac{d}{dx}p\left(x\right)dxdh(x)+dxdp(x)
ddx[h(x)p(x)]=\frac{d}{dx}\left[h\left(x\right)p\left(x\right)\right]=dxd[h(x)p(x)]=
ddx[h(x)p(x)+h(x) ddx [p(x)]]\frac{d}{dx}\left[h\left(x\right)p\left(x\right)+h\left(x\right)\ \frac{d}{dx}\ \left[p\left(x\right)\right]\right]dxd[h(x)p(x)+h(x) dxd [p(x)]]
ddx[h(x)p(x)+h(x) ddx [g(x)]]\frac{d}{dx}\left[h\left(x\right)p\left(x\right)+h\left(x\right)\ \frac{d}{dx}\ \left[g\left(x\right)\right]\right]dxd[h(x)p(x)+h(x) dxd [g(x)]]
It is done.