Chương 5: ĐẠO HÀM - Hỏi đáp

Giúp e với ạ

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:

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

a) \(x^3+2x^2y+xy^2-9x\)

\(=x\left(x^2+2xy+y^2-9\right)\)

\(=x\left[\left(x+y\right)^2-3^2\right]=x\left(x+y+3\right)\left(x+y-3\right)\)

b) \(2x-2y-x^2+2xy-y^2\)

\(=2\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(2-x+y\right)\)

c) \(x^2-2x-4y^2-4y\)

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

\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y-2\right)\)

Lời giải:

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

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

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

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

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

a) y' = 5cosx -3(-sinx) = 5cosx + 3sinx;

b) = = .

c) y' = cotx +x. = cotx -.

d) + = = (x. cosx -sinx).

e) = = .

f) y' = (√(1+x²))' cos√(1+x²) = cos√(1+x²) = cos√(1+x²).

Lời giải:

Ta có:

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

\(y'=0\Leftrightarrow e^x-\frac{1}{e^x}=0\Leftrightarrow e^{2x}-1=0\)

\(\Leftrightarrow e^{2x}=1\Leftrightarrow 2x=0\Leftrightarrow x=0\)

ta có : \(y'=\left(x\left(1+sin2x\right)+\dfrac{1}{2cos2x}\right)'\)

\(=x'\left(1+sin2x\right)+\left(1+sin2x\right)'.x-\dfrac{\left(2x\right)'}{2cos^22x}\)

\(=1+sin2x+2xcos2x-\dfrac{1}{cos^22x}\)

\(\Rightarrow y''=\left(y'\right)'=\left(1+sin2x+2xcos2x-\dfrac{1}{cos^22x}\right)'\)

\(=2cos2x+2cos2x-2xsin2x+\dfrac{2}{cos^42x}\)

\(=4cos2x-2xsin2x+\dfrac{2}{cos^42x}\)

sin \(\sqrt{x^3-2x+5}\)

\(\left(sin\sqrt{x^3-2x+5}\right)'\)

\(\left(x^3-2x+5\right)'cos\sqrt{x^3-2x+5}\)

\(\left(3x^2-2\right)cos\sqrt{x^3-2x+5}\)

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

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

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

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

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

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

a) ta có : \(\left(y\right)'=\left(x^4-3x^3+\sqrt{\dfrac{x-3}{4}}\right)'=\left(x^4-3x^3+\left(\dfrac{x-3}{4}\right)^{0,5}\right)'\)

\(=\left(x^4\right)'-\left(3x^3\right)'+\left(\left(\dfrac{x-3}{4}\right)^{0,5}\right)'=4x^3-9x^2+\dfrac{1}{2}\left(\dfrac{x-3}{4}\right)^{-0,5}\)

\(=4x^3-9x^2+\dfrac{1}{2\sqrt{\dfrac{x-3}{4}}}\)

câu b với câu c ; mk o hiểu cái đề

\(\left[\left(x^7+7x^5\right)\left(x^3+2x^2\right)\right]'\)

\(\left(x^7+7x^5\right)'\left(x^3+2x^2\right)+\left(x^7+7x^5\right)\left(x^3+2x^2\right)'\)

\(\left(7x^6+35x^4\right)\left(x^3+2x^2\right)+\left(x^7+7x^5\right)\left(3x^2+4x\right)\)

\(x^6\left[\left(7x^2+35\right)\left(x+2\right)+\left(x^2+7\right)\left(3x+4\right)\right]\)

\(x^6\left(10x^3+18x^2+56x+98\right)\)