Processing math: 100%
Derivácia funkcie - Úlohy
Derivácia funkcie
Úlohy
- \displaystyle f(x)=2x^6-5x^{-2}-\frac{34}{x^4}+\sin x
- \displaystyle g(x)=2x^2-14x^{-3}-\frac{3}{\sin x}+\arcsin x
- \displaystyle h(x)=\frac{\ln x}{x^4+2x^3}+ 9
- \displaystyle f(x)=\sqrt{x^7}-\frac{\sqrt[4]{x^5}\cdot x^{-2}}{x^4\cdot\sqrt[3]{x^2}}
- \displaystyle g(x)=e^{x}\cdot \sin x-\frac{4}{x}+\arctan x
- \displaystyle h(x)=\frac{\ln x+\cos x- 2x^3}{\sin x}
- \displaystyle f(x)=x^2\cdot e^x \cdot \sin x
- \displaystyle g(x)=\sqrt{x}\cdot \arcsin x-\tan x\cdot (4x-4)
- \displaystyle h(x)=\frac{1-x^2}{x^2+1}
- \displaystyle f(x)=e^x-3^x+\log_3 x
- \displaystyle g(x)=\frac{3e^x\cdot \cos x}{1-x^4+x^3}
- \displaystyle h(x)=(3+\sin x-3x^2)\cdot(3\ln x-\arctan x)
- \displaystyle f(x)=\frac{e^x}{x^3}\cdot\left(1+\frac{\sin x}{x}\right)
- \displaystyle g(x)=\frac{1-x}{1+x}+\frac{1-x^3}{1+x^2}
- \displaystyle h(x)=\sin x\cdot\tan x\cdot \cos x
- \displaystyle f(x)=\sqrt[3]{x}-\frac{3}{\sqrt[4]{x^3}}+x^{-5}
- \displaystyle g(x)=\frac{4x^4+\sin x}{\arccos x}
- \displaystyle h(x)=2^x-5x^{-4}-\frac{\tan x}{x^4}+x^3\cos x
- \displaystyle f(x)=(x^3\cdot e^x+x^3-5)\cdot(11\sin x+e^x\cdot\cos x)
- \displaystyle g(x)=x\cdot\ln x-\sqrt[5]{x^{\frac{3}{2}}}
- \displaystyle h(x)=\frac{1}{2}\arctan{\left(\frac{x}{4}\right)}
- \displaystyle f(x)=\frac{\ln (x^2+1)}{x}+\sqrt{x^4+4x^3+2x^2-2}
- \displaystyle g(x)=\arctan{(2x^4+3x^2+3)}
- \displaystyle h(x)=\ln{\sqrt{x^2+2x+1}}
- \displaystyle f(x)=\arccos{\frac{3x-1}{4}}
- \displaystyle g(x)=\ln{(1+\sin^2 x)}
- \displaystyle h(x)=\frac{x}{2}+\frac{1}{4}\sin(2x)
- \displaystyle f(x)=\ln{(1+\tan x)}
- \displaystyle g(x)=-2\sqrt{1-e^x}
- \displaystyle h(x)=-\frac{\sqrt{x^2+1}}{x-1}
- \displaystyle f(x)=-\frac{x}{\sqrt{5-x^2}}
- \displaystyle g(x)=\ln\left(\frac{x+\sqrt{4+x^2}}{x^3+2x+1}\right)
- \displaystyle h(x)=\sqrt{\frac{x-1}{x+1}}
- \displaystyle f(x)=-2\arctan{\sqrt{\frac{3-x}{x-1}}}
- \displaystyle g(x)=\ln({e^x+\sqrt{e^{2x}-1}})+\arctan{\sqrt{e^{2x}-1}}
- \displaystyle h(x)=\sin\left(\frac{2x^3-5x^2+16x-9}{x\cdot e^{3x-2}}\right)
- \displaystyle f(x)=\frac{x^2\cos(3x^2-6x)}{(3x-5)^4}
- \displaystyle g(x)=\frac{25}{2}\arctan^4{(1+\sin x^2)}
- \displaystyle h(x)=e^{\cos x}\cdot\sin^2(3x^5-6x^3)
- \displaystyle g(x)=x\cdot \sqrt{1+x^2}-\frac{1}{(1+x^2)^3}
Žiadne komentáre:
Zverejnenie komentára