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(m+n)′(x)=m′(x)+n′(x)=−3sinx+4x\left(m+n\right)'\left(x\right)=m'\left(x\right)+n'\left(x\right)=-3\sin x+4x(m+n)′(x)=m′(x)+n′(x)=−3sinx+4x
(m+n)′(x)=m′(x)+n′(x)=3sinx\left(m+n\right)'\left(x\right)=m'\left(x\right)+n'\left(x\right)=3\sin x(m+n)′(x)=m′(x)+n′(x)=3sinx
(m+n)′(x)=m′(x)+n′(x)=3sinx+x\left(m+n\right)'\left(x\right)=m'\left(x\right)+n'\left(x\right)=3\sin x+x(m+n)′(x)=m′(x)+n′(x)=3sinx+x
(a+b)′(x)=a′(x)+b′(x)=12x+3x\left(a+b\right)'\left(x\right)=a'\left(x\right)+b'\left(x\right)=\frac{1}{2\sqrt{x}}+\frac{3}{x}(a+b)′(x)=a′(x)+b′(x)=2x1+x3
(a+b)′(x)=a′(x)+b′(x)=12x\left(a+b\right)'\left(x\right)=a'\left(x\right)+b'\left(x\right)=\frac{1}{2\sqrt{x}}(a+b)′(x)=a′(x)+b′(x)=2x1
(a+b)′(x)=a′(x)+b′(x)=3x\left(a+b\right)'\left(x\right)=a'\left(x\right)+b'\left(x\right)=\frac{3}{x}(a+b)′(x)=a′(x)+b′(x)=x3
(d+e)′(x)=d′(x)+e′(x)=−3x4+104\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=-\frac{3}{x^4}+10^4(d+e)′(x)=d′(x)+e′(x)=−x43+104
(d+e)′(x)=d′(x)+e′(x)=−3x4−104\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=-\frac{3}{x^4}-10^4(d+e)′(x)=d′(x)+e′(x)=−x43−104
(d+e)′(x)=d′(x)+e′(x)=3x4+104\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{3}{x^4}+10^4(d+e)′(x)=d′(x)+e′(x)=x43+104
c′(x)=ex+sinx+1xc'\left(x\right)=e^x+\sin x+\frac{1}{x}c′(x)=ex+sinx+x1
c′(x)=ex+sinxc'\left(x\right)=e^x+\sin xc′(x)=ex+sinx
c′(x)=ex+1xc'\left(x\right)=e^x+\frac{1}{x}c′(x)=ex+x1
(d+e)′(x)=d′(x)+e′(x)=2x3+12x2\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{2}{x^3}+12x^2(d+e)′(x)=d′(x)+e′(x)=x32+12x2
(d+e)′(x)=d′(x)+e′(x)=2x3−12x2\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{2}{x^3}-12x^2(d+e)′(x)=d′(x)+e′(x)=x32−12x2
(d+e)′(x)=d′(x)+e′(x)=2x3+12x\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{2}{x^3}+12x^{ }(d+e)′(x)=d′(x)+e′(x)=x32+12x
Find the derivative of the function h(x)=x−lnxh\left(x\right)=\sqrt{x}-\ln xh(x)=x−lnx .
h′(x)=12x−1xh'\left(x\right)=\frac{1}{2\sqrt{x}}-\frac{1}{x}h′(x)=2x1−x1
h′(x)=12x+1xh'\left(x\right)=\frac{1}{2\sqrt{x}}+\frac{1}{x}h′(x)=2x1+x1
h′(x)=1x+1xh'\left(x\right)=\frac{1}{\sqrt{x}}+\frac{1}{x}h′(x)=x1+x1
r′(x)=1xr'\left(x\right)=\frac{1}{x}r′(x)=x1
r′(x)=1x−cosxr'\left(x\right)=\frac{1}{x-\cos x}r′(x)=x−cosx1
r′(x)=1cosxr'\left(x\right)=\frac{1}{\cos x}r′(x)=cosx1
(f+g)′(x)′=f′(x)+g′(x)=4+ex\left(f+g\right)'\left(x\right)'=f'\left(x\right)+g'\left(x\right)=4+e^x(f+g)′(x)′=f′(x)+g′(x)=4+ex
(f+g)′(x)′=f′(x)+g′(x)=4\left(f+g\right)'\left(x\right)'=f'\left(x\right)+g'\left(x\right)=4(f+g)′(x)′=f′(x)+g′(x)=4
(f+g)′(x)′=f′(x)+g′(x)=ex\left(f+g\right)'\left(x\right)'=f'\left(x\right)+g'\left(x\right)=e^x(f+g)′(x)′=f′(x)+g′(x)=ex
(f+g)′(x)=f′(x)+g′(x)=−sinx+1x\left(f+g\right)'\left(x\right)=f'\left(x\right)+g'\left(x\right)=-\sin x+\frac{1}{x}(f+g)′(x)=f′(x)+g′(x)=−sinx+x1
(f+g)′(x)=f′(x)+g′(x)=−sinx−1x\left(f+g\right)'\left(x\right)=f'\left(x\right)+g'\left(x\right)=-\sin x-\frac{1}{x}(f+g)′(x)=f′(x)+g′(x)=−sinx−x1
(f+g)′(x)=f′(x)−g′(x)=−sinx−1x\left(f+g\right)'\left(x\right)=f'\left(x\right)-g'\left(x\right)=-\sin x-\frac{1}{x}(f+g)′(x)=f′(x)−g′(x)=−sinx−x1
Determine the derivative of q(x)=ex+tanxq\left(x\right)=e^x+\tan xq(x)=ex+tanx .
q′(x)=ex+sec2q'\left(x\right)=e^x+\sec^2q′(x)=ex+sec2
q′(x)=ex+sec2xq'\left(x\right)=e^x+\sec^2xq′(x)=ex+sec2x
q′(x)=ex+secq'\left(x\right)=e^x+\sec^{ }q′(x)=ex+sec
It is done.