Cho hàm số \(f\left(x\right)=\dfrac{\left(\sqrt{x}-1\right)^2}{x}\). Tính giá trị biểu thức \(0,01.f'\left(0,01\right)\). 9 -9 90 -90 Hướng dẫn giải: \(f\left(x\right)=\dfrac{x-2\sqrt{x}+1}{x}=1-2.x^{-\dfrac{1}{2}}+\dfrac{1}{x}\), \(f'\left(x\right)=-2.\left(-\dfrac{1}{2}\right)x^{-\dfrac{3}{2}}-\dfrac{1}{x^2}\Rightarrow f'\left(0,01\right)=\left(10^{-2}\right)^{-\dfrac{3}{2}}-\left(10^{-2}\right)^{-2}=10^3-10^4=-9.10^3\) Do đó \(0,01.f'\left(0,01\right)=10^{-2}.\left(-9.10^3\right)=-90\)
Khẳng định nào sau đây là sai ? \(y=\left(1-\dfrac{2}{x^2}\right)^3\Rightarrow y'=\dfrac{12}{x^3}\left(1-\dfrac{2}{x^2}\right)\) \(y=\frac{x^2}{x^2+1}\Rightarrow y'=\frac{2x}{\left(x^2+1\right)^2}\) \(y=\frac{x}{1-4x}\Rightarrow y'=\frac{1}{\left(1-4x\right)^2}\) \(y=\frac{\sqrt{x}}{\sqrt{x}+1}\Rightarrow y'=\frac{1}{2\sqrt{x}\left(\sqrt{x}+1\right)}\) Hướng dẫn giải: a) \(y=\left(1-\dfrac{2}{x^2}\right)^3\Rightarrow y'=3\left(1-\dfrac{2}{x^2}\right)^2.\left(1-\dfrac{2}{x^2}\right)'=\dfrac{12}{x^3}\left(1-\dfrac{2}{x^2}\right)^2\) b) \(y=\dfrac{x^2}{x^2+1}\Rightarrow y'=\dfrac{2x\left(x^2+1\right)-x^2.2x}{\left(x^2+1\right)^2}=\dfrac{2x}{\left(x^2+1\right)^2}\) c) \(y=\dfrac{x}{1-4x}\Rightarrow y'=\dfrac{1.\left(1-4x\right)-x.\left(-4\right)}{\left(1-4x\right)^2}=\dfrac{1}{\left(1-4x\right)^2}\) d) \(y=\dfrac{\sqrt{x}}{\sqrt{x}+1}=1-\dfrac{1}{\sqrt{x}+1}\Rightarrow y'=\dfrac{\left(\sqrt{x}+1\right)'}{\left(\sqrt{x}+1\right)^2}=\dfrac{1}{2\sqrt{x}\left(\sqrt{x}+1\right)^2}\) Khẳng định sai là \(y=\frac{\sqrt{x}}{\sqrt{x}+1}\Rightarrow y'=\frac{1}{2\sqrt{x}\left(\sqrt{x}+1\right)}\).
Cho hàm số \(f\left(x\right)=\left(1+\dfrac{1}{\sqrt[3]{x}}\right)^3\). Trong các khẳng định sau, khẳng định nào sai? \(f'\left(1\right)=-4\) \(f'\left(-1\right)=0\) \(f'\left(8\right)=-\dfrac{9}{64}\) \(f'\left(-8\right)=-\dfrac{1}{8}\) Hướng dẫn giải: Cách 1: Dùng MTCT. Cách 2: Đặt \(u=1+x^{-\dfrac{1}{3}}\) thì \(f\left(x\right)=u^3\), do đó \(f'\left(x\right)=3u^2.u'=3\left(1+\dfrac{1}{\sqrt[3]{x}}\right)^2.\left(-\dfrac{1}{3}x^{-\dfrac{4}{3}}\right)=-\dfrac{1}{x^{\dfrac{4}{3}}}\left(1+\dfrac{1}{\sqrt[3]{x}}\right)^2\) nên \(f'\left(1\right)=-4,f'\left(-1\right)=0,f'\left(8\right)=-\dfrac{9}{64},f'\left(-8\right)=-\dfrac{1}{64}\)
Cho hàm số \(y=f\left(x\right)=\sqrt{x+2\sqrt{x}}\) Tính \(f'\left(1\right)\) . \(\frac{\sqrt{2}}{2}\) \(\sqrt{2}\) \(\frac{\sqrt{3}}{3}\) \(\sqrt{3}\) Hướng dẫn giải: Cách 1: Dùng MTCT. Cách 2:
Tính đạo hàm của hàm số \(y=\dfrac{x^2-x+1}{x^2+x+1}\) \(y'=\dfrac{2\left(x^2-1\right)}{\left(x^2+x+1\right)}\) \(y'=\dfrac{2\left(x^2-1\right)}{\left(x^2+x+1\right)^2}\) \(y'=\dfrac{-2\left(x^2-1\right)}{\left(x^2+x+1\right)^2}\) \(y'=\dfrac{2\left(x^2+1\right)}{\left(x^2+x+1\right)^2}\) Hướng dẫn giải: Sử dụng công thức \(\left(\dfrac{ax^2+bx+c}{mx^2+nx+p}\right)'=\dfrac{\left|\begin{matrix}a&b\\m&n\end{matrix}\right|x^2+2\left|\begin{matrix}a&c\\m&n\end{matrix}\right|x+\left|\begin{matrix}b&c\\n&p\end{matrix}\right|}{\left(mx^2+nx+p\right)^2}\). Đáp số : \(y'=\dfrac{2\left(x^2-1\right)}{\left(x^2+x+1\right)^2}\)
Tính đạo hàm của hàm số \(y=\dfrac{x}{\left(1-x^2\right)\left(1+x\right)^3}\) \(y'=\dfrac{4x^2-x+1}{\left(1-x\right)^3\left(1+x\right)^4}\) \(y'=\dfrac{4x^2+3x+1}{\left(1-x^2\right)^2\left(1+x\right)^3}\) \(y'=\dfrac{4x^2-3x+1}{\left(1-x\right)^2\left(1+x\right)^5}\) \(y'=\dfrac{4x^2+x+1}{\left(1-x\right)^3\left(1+x\right)^4}\) Hướng dẫn giải: \(y=\dfrac{x}{\left(1-x^2\right)\left(1+x\right)^3}=\dfrac{u}{v}\) với \(u=x,v=\left(1-x^2\right)\left(1+x\right)^3\). Theo quy tắc đạo hàm một thương thì \(y'=\dfrac{u'v-uv'}{v^2}\) trong đó \(u'=1,v'=-2x\left(1+x\right)^3+\left(1-x^2\right)3\left(1+x\right)^2=\left(1+x\right)^3\left[-2x+3\left(1-x\right)\right]=\left(1+x\right)^3\left(-5x+3\right)\) \(u'v-uv'=1.\left(1-x^2\right)\left(1+x\right)^3-x.\left(1+x\right)^3\left(-5x+3\right)=\left(1+x\right)^3\left[\left(1-x^2\right)-x\left(-5x+3\right)\right]=\left(1+x\right)^3\left(4x^2-3x+1\right)\) \(y'=\dfrac{\left(1+x\right)^3\left(4x^2-3x+1\right)}{\left(1-x^2\right)^2\left(1+x\right)^6}=\dfrac{4x^2-3x+1}{\left(1-x^2\right)^2\left(1+x\right)^3}=\dfrac{4x^2-3x+1}{\left(1-x\right)^2\left(1+x\right)^5}\)
Tính đạo hàm của hàm số \(y=\dfrac{x}{\left(1-x\right)^2\left(1+x\right)^3}\) \(y'=\dfrac{4x^2-x+1}{\left(1-x\right)^3\left(1+x\right)}\) \(y'=\dfrac{4x^2-x+1}{\left(1-x\right)\left(1+x\right)^4}\) \(y'=\dfrac{4x^2-x+1}{\left(1-x\right)^3\left(1+x\right)^4}\) \(y'=\dfrac{4x^2-x+1}{\left(1-x\right)^2\left(1+x\right)^4}\) Hướng dẫn giải: \(y=\dfrac{x}{\left(1-x\right)^2\left(1+x\right)^3}=\dfrac{u}{v}\), trong đó \(u=x,v=\left(1-x\right)^2\left(1+x\right)^3\) . Theo quy tắc đạo hàm một thương: \(y'=\dfrac{u'v-uv'}{v^2}\). Ta có \(u'=1,v'=-2\left(1-x\right)\left(1+x\right)^3+\left(1-x\right)^23\left(1+x\right)^2=\left(1-x\right)\left(1+x\right)^2\left[-2\left(1+x\right)+3\left(1-x\right)\right]=\left(1-x\right)\left(1+x\right)^2\left(-5x+1\right)\). \(u'v-uv'=1.\left(1-x\right)^2\left(1+x\right)^3-x\left(1-x\right)\left(1+x\right)^2\left(-5x+1\right)=\left(1-x\right)\left(1+x\right)^2\left[\left(1-x^2\right)-x\left(-5x+1\right)\right]=\left(1-x\right)\left(1+x\right)^2\left(4x^2-x+1\right)\) \(y'=\dfrac{\left(1-x\right)\left(1+x\right)^2\left(4x^2-x+1\right)}{\left(1-x\right)^4\left(1+x\right)^6}=\dfrac{4x^2-x+1}{\left(1-x\right)^3\left(1+x\right)^4}\)
Tính đạo hàm của hàm số \(y=\dfrac{\left(1-x\right)^p}{\left(1+x\right)^q}\). \(y'=\dfrac{\left(1-x\right)^{p-1}\left[\left(p-q\right)x+p+q\right]}{\left(1+x\right)^{q+1}}\) \(y'=\dfrac{-\left(1-x\right)^{p-1}\left[\left(p-q\right)x+p+q\right]}{\left(1+x\right)^{q+1}}\) \(y'=\dfrac{-\left(1-x\right)^{p-1}\left[\left(p+q\right)x+p-q\right]}{\left(1+x\right)^{q+1}}\) \(y'=\dfrac{-\left(1-x\right)^{p-1}\left[\left(p-q\right)x-p+q\right]}{\left(1+x\right)^{q+1}}\) Hướng dẫn giải: \(y=\dfrac{\left(1-x\right)^p}{\left(1+x\right)^q}=\dfrac{u}{v}\) với \(u=\left(1-x\right)^p,v=\left(1+x\right)^q\). Áp dụng quy tắc đạo hàm một thương \(y'=\dfrac{u'v-uv'}{v^2}\). Ta có \(u'=-p\left(1-x\right)^{p-1},v'=q\left(1+x\right)^q\) \(u'v-uv'=-p\left(1-x\right)^{p-1}\left(1+x\right)^q-\left(1-x\right)^pq\left(1+x\right)^{q-1}=\left(1-x\right)^{p-1}\left(1+x\right)^q\left[-p\left(1+x\right)-q\left(1-x\right)\right]=-\left(1-x\right)^{p-1}\left(1+x\right)^{q-1}[\left(p-q\right)x+p+q]\) Từ đó \(y'=\dfrac{-\left(1-x\right)^{p-1}\left(1+x\right)^{q-1}\left[\left(p-q\right)x+p+q\right]}{\left(1+x\right)^{2q}}=\dfrac{-\left(1-x\right)^{p-1}\left[\left(p-q\right)x+p+q\right]}{\left(1+x\right)^{q+1}}\)
