Bài 3: Đạo hàm của hàm số lượng giác

Bài 8:

Ta có: \(\lim\limits_{x\rightarrow2^+}f\left(x\right)=\lim\limits_{x\rightarrow2^+}\dfrac{x^3-8}{2x^2-x-6}=\lim\limits_{x\rightarrow2^+}\dfrac{x^2+2x+4}{2x+3}=\dfrac{12}{7}\)

\(\lim\limits_{x\rightarrow2^-}f\left(x\right)=\lim\limits_{x\rightarrow2^-}\left(mx+10\right)=2m+10\)

\(f\left(2\right)=2m+10\)

Để f(x) liên tục tại x₀ = 2 thì \(\lim\limits_{x\rightarrow2^+}f\left(x\right)=\lim\limits_{x\rightarrow2^-}f\left(x\right)=f\left(2\right)\)

\(\Rightarrow2m+10=\dfrac{12}{7}\Rightarrow m=\dfrac{-29}{7}\)

Vậy...

Bài 9: 

Ta có: \(\lim\limits_{x\rightarrow0}f\left(x\right)=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+x}-\sqrt{1-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{1+x-\left(1-x\right)}{x\left(\sqrt{1+x}+\sqrt{1-x}\right)}=\lim\limits_{x\rightarrow0}\dfrac{2}{\sqrt{1+x}+\sqrt{1-x}}=1\)

\(f\left(0\right)=-5m+2\)

Để f(x) liên tục tại x₀ = 0 thì \(\lim\limits_{x\rightarrow0}f\left(x\right)=f\left(0\right)\)

\(\Rightarrow-5m+2=1\Rightarrow m=\dfrac{1}{5}\)

Vậy...

Câu 10:

Ta có: 

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{2x-1}-1}{x^2+2x-3}=\lim\limits_{x\rightarrow1^+}\dfrac{2x-1-1}{\left(x-1\right)\left(x+3\right)\left(\sqrt{2x-1}+1\right)}\)

\(=\lim\limits_{x\rightarrow1^+}\dfrac{2}{\left(x+3\right)\left(\sqrt{2x-1}+1\right)}=\dfrac{1}{4}\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(x+m\right)=m+1\)

\(f\left(1\right)=m+1\)

Để f(x) liên tục tại x₀ = 1 thì \(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=f\left(1\right)\)

\(\Rightarrow m+1=\dfrac{1}{4}\Rightarrow m=\dfrac{-3}{4}\)

Vậy...

Bài 11:

Ta có: \(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{\sqrt[3]{6+x}-2}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{6+x-8}{\left(x-2\right)\left(\sqrt[3]{\left(6+x\right)^2}+2\sqrt[3]{6+x}+4\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\sqrt[3]{\left(6+x\right)^2}+2\sqrt[3]{6+x}+4}=\dfrac{1}{12}\)

\(f\left(2\right)=4-m\)

Để f(x) liên tục tại x₀ = 2 thì \(\lim\limits_{x\rightarrow2}f\left(x\right)=f\left(2\right)\)

\(\Rightarrow4-m=\dfrac{1}{12}\Rightarrow m=\dfrac{47}{12}\)

Vậy...

Bài 12:

Ta có: \(\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{12x-4}-2}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{12x-4-8}{\left(x-1\right)\left(\sqrt[3]{\left(12x-4\right)^2}+2\sqrt[3]{12x-4}+4\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{12}{\sqrt[3]{\left(12x-4\right)^2}+2\sqrt[3]{12x-4}+4}=1\)

\(f\left(1\right)=\sqrt{m^2+8}+2m\)

Để f(x) liên tục thì \(\lim\limits_{x\rightarrow1}f\left(x\right)=f\left(1\right)\)

\(\Rightarrow\sqrt{m^2+8}+2m=1\)

\(\Leftrightarrow\sqrt{m^2+8}=1-2m\)

\(\Rightarrow m^2+8=1-4m+4m^2\) (ĐK: \(1-2m\ge0\Rightarrow m\le\dfrac{1}{2}\))

\(\Leftrightarrow3m^2-4m-7=0\)

\(\Rightarrow\left[{}\begin{matrix}m=\dfrac{7}{3}\left(ktm\right)\\m=-1\left(tm\right)\end{matrix}\right.\)

Vậy...

\(y'=2cos\left(x^2+2\right).\left[cos\left(x^2+2\right)\right]'=-4xcos\left(x^2+2\right).sin\left(x^2+2\right)\)

\(=-4x.sin\left(2x^2+4\right)\)

\(y=\dfrac{x^2+3x+3}{x^2+1}\Rightarrow y'=\dfrac{\left(x^2+3x+3\right)'\left(x^2+1\right)-\left(x^2+3x+3\right)\left(x^2+1\right)'}{\left(x^2+1\right)^2}\)

\(y'=\dfrac{\left(x^2+1\right)\left(2x+3\right)-\left(x^2+3x+3\right).2x}{\left(x^2+1\right)^2}\)

\(y'=0\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)-2x\left(x^2+3x+3\right)=0\)

\(\Leftrightarrow2x^3+3x^2+2x+3-2x^3-6x^2-6x=0\)

\(\Leftrightarrow3x^2+4x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=..\\x=...\end{matrix}\right.\)

Check lai ho t nhe

\(\lim\limits_{x\rightarrow0}\dfrac{2\left(\sqrt{3x+1}-1\right)}{x}=\lim\limits_{x\rightarrow0}\dfrac{6x}{x\left(\sqrt{3x+1}+1\right)}=\lim\limits_{x\rightarrow0}\dfrac{6}{\sqrt{3x+1}+1}=3\)

\(\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x-2\right)}{x+1}=\lim\limits_{x\rightarrow-1}\left(x-2\right)=-3\)

\(\Rightarrow I-J=6\)

Lời giải:

Em không rõ ở phần tìm đạo hàm theo định nghĩa (lim) hay tìm đạo hàm dựa theo công thức

Thông thường lớp 11 thì thường áp dụng luôn công thức

Áp dụng công thức: \((u^{\alpha})'=\alpha.u'.u^{\alpha-1}\) thì:

\(y=(x+\sqrt{1+x^2})^{\frac{1}{2}}\)

\(\Rightarrow y'=\frac{1}{2}(x+\sqrt{x^2+1})'(x+\sqrt{x^2+1})^{\frac{1}{2}-1}\)

\(=\frac{(x+\sqrt{x^2+1})'}{2\sqrt{x+\sqrt{x^2+1}}}(*)\)

\((x+\sqrt{x^2+1})'=x'+(\sqrt{x^2+1})'=1+((x^2+1)^{\frac{1}{2}})'\)

\(=1+\frac{1}{2}(x^2+1)'(x^2+1)^{\frac{1}{2}-1}\)

\(=1+\frac{1}{2}.2x.\frac{1}{\sqrt{x^2+1}}=1+\frac{x}{\sqrt{x^2+1}}(**)\)

Từ \((*);(**)\Rightarrow y'=\frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}.2\sqrt{x+\sqrt{x^2+1}}}=\frac{1}{2}\sqrt{\frac{x+\sqrt{x^2+1}}{x^2+1}}\)

ta có : \(y'=\left(\sqrt{x+\sqrt{1+x^2}}\right)'=\dfrac{1}{2\sqrt{x+\sqrt{1+x^2}}}\left(x+\sqrt{1+x^2}\right)'\)

Lời giải:

\(y=\cot ^2x+\cot 2x\)

\(\Rightarrow y'=(\cot ^2x)'+(\cot 2x)'\)

\(=2(\cot x)'\cot x+\frac{-(2x)'}{\sin ^22x}\)

\(=2.\frac{-1}{\sin ^2x}\cot x-\frac{2}{\sin ^22x}\)

\(=-2(\frac{\cot x}{\sin ^2x}+\frac{1}{\sin ^22x})\)

\(y'=\left(cos3x\times sin2x\right)'\)

\(\left(cos3x\right)'sin2x+cos3x\left(sin2x\right)'\)

\(-\left(3x\right)'sin3x\sin2x+\left(2x\right)'\cos2x\cos3x\)

\(-3\sin3x\sin2x+2\cos2x\cos3x\)

\(\dfrac{-3}{2}\left[\cos x-\cos5x\right]+\left[\cos x+\cos5x\right]\)

\(\dfrac{5}{2}\cos5x-\dfrac{1}{2}\cos x\)

Lời giải:

Ta có:

\(f(x)=\sin ^2\left(\frac{\pi}{6}-x\right)+\sin ^2\left(\frac{\pi}{6}+x\right)\)

\(\Rightarrow f'(x)=2\sin \left(\frac{\pi}{6}-x\right).-\cos \left(\frac{\pi}{6}-x\right)+2\sin \left(\frac{\pi}{6}+x\right)\cos \left(\frac{\pi}{6}+x\right)\)

\(f'(x)=-\sin 2\left(\frac{\pi}{6}-x\right)+\sin 2\left(\frac{\pi}{6}+x\right)\)

Áp dụng công thức: \(\sin a-\sin b=2\cos \frac{a+b}{2}\sin \frac{a-b}{2}\) suy ra:

\(f'(x)=-\sin \left(\frac{\pi}{3}-2x\right)+\sin \left(\frac{\pi}{3}+2x\right)\)

\(f'(x)=2\cos \left(\frac{\pi}{3}\right)\sin 2x=\sin 2x\) (đpcm)

c

cos x