Tính đạo hàm của hàm số \(y=\dfrac{\left(2-x^2\right)\left(3-x^3\right)}{\left(1-x\right)^2}\) \(y'=\dfrac{-3x^5+5x^4+2x^3-6x^2-6x+12}{\left(1-x\right)^4}\) \(y'=\dfrac{-3x^5+5x^4+2x^3-6x^2-6x+12}{\left(1-x\right)^3}\) \(y'=\dfrac{-3x^5-5x^4+2x^3+6x^2-6x+12}{\left(1-x\right)^3}\) \(y'=\dfrac{-3x^5+x^4+2x^3-3x^2-6x+12}{\left(1-x\right)^3}\) Hướng dẫn giải: \(y=\dfrac{\left(2-x^2\right)\left(3-x^3\right)}{\left(1-x\right)^2}=\dfrac{x^5-2x^3-3x^2+6}{\left(1-x\right)^2}=\dfrac{u}{v}\) với \(u=x^5-2x^3-3x^2+6,v=\left(1-x\right)^2\). Ta có \(y'=\dfrac{u'v-uv'}{v^2}\). \(u'=5x^4-6x^2-6x,v'=-2\left(1-x\right)\) ; \(v^2=\left(1-x\right)^4\) \(u'v-uv'=\left(5x^4-6x^2-6x\right)\left(1-x\right)^2+2\left(x^5-2x^3-3x^2+6\right)\left(1-x\right)=\left(1-x\right)\left[\left(5x^4-6x^2-6x\right)\left(1-x\right)+2\left(x^5-2x^3-3x^2+6\right)\right]\) \(=\left(1-x\right)\left(-3x^5+5x^4+2x^3-6x^2-6x+12\right)\) . Do đó \(y'=\dfrac{-3x^5+5x^4+2x^3-6x^2-6x+12}{\left(1-x\right)^3}\)
Tính đạo hàm của hàm số \(y=\sqrt[m+n]{\left(1-x\right)^m\left(1+x\right)^n}\) . \(\dfrac{\left(m+n\right)+\left(m+n\right)x}{\left(m+n\right)\sqrt[m+n]{\left(1-x\right)^n\left(1+x\right)^m}}\) \(\dfrac{\left(m+n\right)-\left(m-n\right)x}{\left(m+n\right)\sqrt[m+n]{\left(1-x\right)^n\left(1+x\right)^m}}\) \(\dfrac{\left(m-n\right)-\left(m+n\right)x}{\left(m+n\right)\sqrt[m+n]{\left(1-x\right)^n\left(1+x\right)^m}}\) \(\dfrac{\left(m+n\right)+\left(m-n\right)x}{\sqrt[m+n]{\left(1-x\right)^n\left(1+x\right)^m}}\) Hướng dẫn giải: Ta có \(y=\left(1-x\right)^{\dfrac{m}{m+n}}\left(1+x\right)^{\dfrac{n}{m+n}}\Rightarrow y'=\dfrac{-m}{m+n}\left(1-x\right)^{\dfrac{m}{m+n}-1}\left(1+x\right)^{\dfrac{n}{m+n}}+\left(1-x\right)^{\dfrac{m}{m+n}}.\dfrac{n}{m+n}\left(1+x\right)^{\dfrac{n}{m+n}-1}\) \(y'=-\dfrac{m}{m+n}\left(1-x\right)^{-\dfrac{n}{m+n}}\left(1+x\right)^{\dfrac{n}{m+n}}+\dfrac{n}{m+n}\left(1-x\right)^{\dfrac{m}{m+n}}\left(1+x\right)^{-\dfrac{m}{m+n}}\) \(=-\dfrac{m}{m+n}\left(\dfrac{1+x}{1-x}\right)^{\dfrac{n}{m+n}}+\dfrac{n}{m+n}\left(\dfrac{1-x}{1+x}\right)^{\dfrac{m}{m+n}}\)\(=-\dfrac{m\sqrt[m+n]{\left(1+x\right)^n}}{\left(m+n\right)\sqrt[m+n]{\left(1-x\right)^n}}+\dfrac{n\sqrt[m+n]{\left(1-x\right)^m}}{\left(m+n\right)\sqrt[m+n]{\left(1+x\right)^m}}\) \(=\dfrac{-m\left(1+x\right)+n\left(1-x\right)}{\left(m+n\right)\sqrt[m+n]{\left(1-x\right)^n\left(1+x\right)^m}}=\dfrac{\left(n-m\right)-\left(m+n\right)x}{\left(m+n\right)\sqrt[m+n]{\left(1-x\right)^n\left(1+x\right)^m}}\)
Cho hàm số \(y=f\left(x\right)=\frac{1}{2}ln\left(1+x\right)-\frac{1}{4}ln\left(1+x^2\right)-\frac{1}{2\left(1+x\right)}\) Tính \(f'\left(1\right)\). \(\frac{1}{2}\) \(\frac{1}{4}\) \(\frac{1}{8}\) \(\frac{1}{12}\) Hướng dẫn giải: \(f\left(x\right)=\dfrac{1}{2}\ln\left(1+x\right)-\dfrac{1}{4}\ln\left(1+x^2\right)-\dfrac{1}{2\left(x+1\right)}\), \(f'\left(x\right)=\dfrac{1}{2}.\dfrac{1}{1+x}-\dfrac{1}{4}.\dfrac{2x}{1+x^2}+\dfrac{1}{2\left(1+x\right)^2}=\dfrac{1}{2}\left(\dfrac{1}{1+x}-\dfrac{1}{1+x^2}+\dfrac{1}{\left(1+x\right)^2}\right)\) \(\Rightarrow\) \(f'\left(1\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4}\right)=\dfrac{1}{8}\)
Tìm đạo hàm của hàm số \(y=\frac{1}{4\left(1+x^4\right)}+\frac{1}{4}ln\left(\frac{x^4}{1+x^4}\right)\) . \(\dfrac{1}{x\left(1+x^4\right)^2}\) \(\dfrac{1}{x^2\left(1+x^4\right)^2}\) \(\dfrac{1}{x^3\left(1+x^4\right)^2}\) \(\dfrac{1}{x^4\left(1+x^4\right)}\) Hướng dẫn giải: Ta có \(\left(\dfrac{1}{\left(1+x^4\right)}\right)'=\dfrac{-4x^3}{\left(1+x^4\right)^2}=-\dfrac{4x^3}{\left(1+x^4\right)^2}\Rightarrow\left(\dfrac{1}{4\left(1+x^4\right)}\right)'=-\dfrac{x^3}{\left(1+x^4\right)^2}\) và \(\dfrac{x^4}{1+x^4}=1-\dfrac{1}{1+x^4}\Rightarrow\left(\dfrac{x^4}{1+x^4}\right)'=\left(-\dfrac{1}{1+x^4}\right)'=\dfrac{4x^3}{\left(1+x^4\right)^2}\) suy ra \(\left(\ln\left(\dfrac{x^4}{1+x^4}\right)\right)'=\left(\dfrac{x^4}{1+x^4}\right)':\left(\dfrac{x^4}{1+x^4}\right)=\dfrac{4x^3}{\left(1+x^4\right)^2}.\dfrac{1+x^4}{x^4}=\dfrac{4}{\left(1+x^4\right)x}\)\(\Rightarrow\left(\dfrac{1}{4}\ln\left(\dfrac{x^4}{1+x^4}\right)\right)'=\dfrac{1}{x\left(1+x^4\right)}\) Do đó \(y'=-\dfrac{x^3}{\left(1+x^4\right)^2}+\dfrac{1}{x\left(1+x^4\right)}=\dfrac{-x^4+\left(1+x^4\right)}{x\left(1+x^4\right)^2}=\dfrac{1}{x\left(1+x^4\right)^2}\)
Tính đạo hàm của hàm số \(y=ln\left(\sqrt{1+e^x}-1\right)-ln\left(\sqrt{1+e^x}+1\right)\) . \(\sqrt{1+e^x}\) \(\frac{1}{\sqrt{1+e^x}}\) \(\frac{e^x}{\sqrt{1+e^x}}\) \(\frac{\sqrt{1+e^x}}{e^x}\) Hướng dẫn giải: Áp dụng công thức \(\left(\ln u\right)=\dfrac{u'}{u}\), ta suy ra \(\left(\ln\left(\sqrt{1+e^x}-1\right)\right)'=\dfrac{\dfrac{e^x}{2\sqrt{1+e^x}}}{\sqrt{1+e^x}-1}=\dfrac{e^x}{2\sqrt{1+e^x}.\left(\sqrt{1+e^x}-1\right)}\) \(\left(\ln\left(\sqrt{1+e^x}+1\right)\right)'=\dfrac{\dfrac{e^x}{2\sqrt{1+e^x}}}{\sqrt{1+e^x}+1}=\dfrac{e^x}{2\sqrt{1+e^x}.\left(\sqrt{1+e^x}+1\right)}\) \(y'=\dfrac{e^x}{2\sqrt{1+e^x}\left(\sqrt{1+e^x}-1\right)}-\dfrac{e^x}{2\sqrt{1+e^x}\left(\sqrt{1+e^x}+1\right)}=\dfrac{2e^x}{2\sqrt{1+e^x}.e^x}=\dfrac{1}{\sqrt{1+e^x}}\)
Cho hàm số \(y=f\left(x\right)=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}ln\left(x+\sqrt{x^2-a^2}\right)\). Tính \(f'\left(2a\right)\). \(a\sqrt{2}\) \(a\) \(a\sqrt{3}\) \(2a\) Hướng dẫn giải: Ta có \(\left(\sqrt{x^2-a^2}\right)'=\dfrac{x}{\sqrt{x^2-a^2}}\), \(\left(x+\sqrt{x^2-a^2}\right)'=1+\dfrac{x}{\sqrt{x^2-a^2}}=\dfrac{x+\sqrt{x^2-a^2}}{\sqrt{x^2-a^2}}\). Do đó \(\left(\dfrac{x}{2}\sqrt{x^2-a^2}\right)'=\dfrac{1}{2}\sqrt{x^2-a^2}+\dfrac{x}{2}.\dfrac{x}{\sqrt{x^2-a^2}}=\dfrac{2x^2-a^2}{2\sqrt{x^2-a^2}}\) và \(\left(\ln\left(x+\sqrt{x^2-a^2}\right)\right)'=\dfrac{x+\sqrt{x^2-a^2}}{\sqrt{x^2-a^2}}:\left(x+\sqrt{x^2-a^2}\right)=\dfrac{1}{\sqrt{x^2-a^2}}\) \(y'=\dfrac{2x^2-a^2}{2\sqrt{x^2-a^2}}-\dfrac{a^2}{2\sqrt{x^2-a^2}}=\sqrt{x^2-a^2}\). Vì vậy \(f'\left(2a\right)=\sqrt{4a^2-a^2}=a\sqrt{3}\)
Cho hàm số \(y=\varphi\left(x\right)=\sqrt{x^2+1}-ln\left(\frac{1+\sqrt{x^2+1}}{x}\right)\). Tính \(\varphi'\left(2\right)\) . \(\frac{\sqrt{3}}{2}\) 1 \(\frac{\sqrt{5}}{2}\) \(\frac{\sqrt{6}}{2}\) Hướng dẫn giải: Ta có \(\left(1+\sqrt{x^2+1}\right)'=\left(\sqrt{x^2+1}\right)'=\dfrac{x}{\sqrt{x^2+1}},\) Vì \(\varphi\left(x\right)=\sqrt{x^2+1}-\ln\left(1+\sqrt{x^2+1}\right)+\ln x\Rightarrow\varphi'\left(x\right)=\dfrac{x}{\sqrt{x^2+1}}-\dfrac{x}{\sqrt{x^2+1}}:\left(1+\sqrt{x^2+1}\right)+\dfrac{1}{x}\) Do đó \(\varphi'\left(2\right)=\dfrac{2}{\sqrt{5}}-\dfrac{2}{\sqrt{5}\left(1+\sqrt{5}\right)}+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\left(1-\dfrac{1}{1+\sqrt{5}}\right)+\dfrac{1}{2}=\dfrac{2}{1+\sqrt{5}}+\dfrac{1}{2}=\dfrac{5+\sqrt{5}}{2\left(1+\sqrt{5}\right)}=\dfrac{\sqrt{5}}{2}\)
Cho hàm số \(y=x.e^{-\frac{x^2}{2}}\). Hệ thức nào trong các hệ thức sau là đúng? \(x.y=\left(1+x^2\right).y'\) \(x.y'=\left(1+x^2\right).y\) \(x.y=\left(1-x^2\right).y'\) \(x.y'=\left(1-x^2\right).y\) Hướng dẫn giải: Ta có \(\left(e^{-\dfrac{x^2}{2}}\right)'=-x.e^{-\dfrac{x^2}{2}}\) nên \(y'=1.e^{-\dfrac{x^2}{2}}+x.\left(-xe^{-\dfrac{x^2}{2}}\right)=\left(1-x^2\right)e^{-\dfrac{x^2}{2}}\) suy ra \(xy'=\left(1-x^2\right)xe^{-\dfrac{x^2}{2}}\) hay \(xy'=\left(1-x^2\right)y\) . Đáp số: \(xy'=\left(1-x^2\right)y\)
Cho hàm số \(y=\frac{1}{1+x+\ln x}\). Khẳng định nào trong các khẳng định sau đây đúng? \(xy=y'\left(y\ln x+1\right)\) \(xy'=y\left(y\ln x-1\right)\) \(xy=y\left(y'\ln x-1\right)\) \(xy'=y\left(y\ln x+1\right)\) Hướng dẫn giải: - Ta có \(\left(1+x+\ln x\right)'=1+\dfrac{1}{x}=\dfrac{x+1}{x}\) nên \(y'=\dfrac{-\left(x+1\right)}{x\left(1+x+\ln x\right)^2}\) \(\Rightarrow xy'=\dfrac{-\left(x+1\right)}{\left(1+x+\ln x\right)^2}=\dfrac{1}{1+x+\ln x}.\dfrac{-\left(1+x\right)}{1+x+\ln x}=y.\dfrac{\ln x-\left(1+x+\ln x\right)}{1+x+\ln x}\) \(=\dfrac{1}{1+x+\ln x}\left(\dfrac{1}{1+x+\ln x}.\ln x-1\right)=y\left(y\ln x-1\right)\). Đáp số: \(xy'=y\left(y\ln x-1\right)\)
Cho hàm số \(y=f\left(x\right)=x^2-\dfrac{1}{2x^2}.\) Tính \(f'\left(2\right)-f'\left(-2\right).\) 0 8 \(\frac{33}{4}\) -1 Hướng dẫn giải: Cách 1: Dùng MTCT. Cách 2: \(f\left(x\right)=x^2-\dfrac{1}{2x^2}\Rightarrow f'\left(x\right)=2x+\dfrac{1}{x^3}\). Do đó \(f'\left(2\right)=4+\dfrac{1}{8},f'\left(-2\right)=-4-\dfrac{1}{8}\Rightarrow f'\left(2\right)-f'\left(-2\right)=8+\dfrac{1}{4}=\dfrac{33}{4}\)
