Bài 2: Quy tắc tính đạo hàm

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

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

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

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

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

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

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

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

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

\(f'\left(x\right)=\dfrac{\left(4-3x\right)'}{2\sqrt{4-3x}}=\dfrac{-3}{2\sqrt{4-3x}}\)

\(\Rightarrow f'\left(-4\right)=\dfrac{-3}{2\sqrt{4-3.\left(-4\right)}}=-\dfrac{3}{8}\)

1.

\(f'\left(x\right)=3x^2-6mx+3\left(2m-1\right)\)

\(f'\left(x\right)-6x=3x^2-3.2\left(m+1\right)x+3\left(2m-1\right)>0\)

\(\Leftrightarrow x^2-2\left(m+1\right)x+2m-1>0\)

\(\Leftrightarrow x^2-2x-1>2m\left(x-1\right)\)

Do \(x>2\Rightarrow x-1>0\) nên BPT tương đương:

\(\dfrac{x^2-2x-1}{x-1}>2m\Leftrightarrow\dfrac{\left(x-1\right)^2-2}{x-1}>2m\)

Đặt \(t=x-1>1\Rightarrow\dfrac{t^2-2}{t}>2m\Leftrightarrow f\left(t\right)=t-\dfrac{2}{t}>2m\)

Xét hàm \(f\left(t\right)\) với \(t>1\) : \(f'\left(t\right)=1+\dfrac{2}{t^2}>0\) ; \(\forall t\Rightarrow f\left(t\right)\) đồng biến

\(\Rightarrow f\left(t\right)>f\left(1\right)=-1\Rightarrow\) BPT đúng với mọi \(t>1\) khi \(2m< -1\Rightarrow m< -\dfrac{1}{2}\)

2.

Thay \(x=0\) vào giả thiết:

\(f^3\left(2\right)-2f^2\left(2\right)=0\Leftrightarrow f^2\left(2\right)\left[f\left(2\right)-2\right]=0\Rightarrow\left[{}\begin{matrix}f\left(2\right)=0\\f\left(2\right)=2\end{matrix}\right.\)

Đạo hàm 2 vế giả thiết:

\(-3f^2\left(2-x\right).f'\left(2-x\right)-12f\left(2+3x\right).f'\left(2+3x\right)+2x.g\left(x\right)+x^2.g'\left(x\right)+36=0\) (1)

Thế \(x=0\) vào (1) ta được:

\(-3f^2\left(2\right).f'\left(2\right)-12f\left(2\right).f'\left(2\right)+36=0\)

\(\Leftrightarrow f^2\left(2\right).f'\left(2\right)+4f\left(2\right).f'\left(2\right)-12=0\) (2)

Với \(f\left(2\right)=0\)  thế vào (2) \(\Rightarrow-12=0\) ko thỏa mãn (loại)

\(\Rightarrow f\left(2\right)=2\)

Thế vào (2):

\(4f'\left(2\right)+8f'\left(2\right)-12=0\Leftrightarrow f'\left(2\right)=1\)

\(\Rightarrow A=3.2+4.1\)

3.

Đặt \(g\left(x\right)=f\left(x\right)-f\left(2x\right)\)

\(\Rightarrow g'\left(x\right)=f'\left(x\right)-2f'\left(2x\right)\) 

Thay \(x=1\Rightarrow18=f'\left(1\right)-2f'\left(2\right)\) (1)

Thay \(x=2\Rightarrow2000=f'\left(2\right)-2f'\left(4\right)\Rightarrow4000=2f'\left(2\right)-4f'\left(4\right)\) (2)

Cộng vế (1) và (2):

\(f'\left(1\right)-4f'\left(4\right)=4018\)

Đặt \(h\left(x\right)=f\left(x\right)-f\left(4x\right)\Rightarrow h'\left(x\right)=f'\left(x\right)-4f'\left(4x\right)\)

Thay \(x=1\Rightarrow h'\left(1\right)=f'\left(1\right)-4f'\left(4\right)=4018\)

Lời giải:
a) $y=\frac{3}{x^2}-(x)^{\frac{1}{2}}+\frac{2}{3}(x)^{\frac{3}{2}}$

$y'=\frac{-3(x^2)'}{x^4}-\frac{1}{2}x^{\frac{-1}{2}}+\frac{2}{3}.\frac{3}{2}.x^{\frac{1}{2}}$

$=\frac{-6}{x^3}-\frac{1}{2\sqrt{x}}+x$

b) 

$y=(x+1)^{\frac{1}{2}}[(x)^{\frac{-1}{2}}-1)$

$y'=\frac{1}{2}(x+1)^{\frac{-1}{2}}[(x)^{\frac{-1}{2}}-1)+(x+1)^{\frac{1}{2}}[\frac{-1}{2}x^{-\frac{3}{2}}]=\frac{1}{2\sqrt{x+1}}(\frac{1}{\sqrt{x}}-1)-\frac{1}{2}\sqrt{x+1}.\frac{1}{\sqrt{x^3}}$

c)

\(y'=\frac{[(x+1)^2]'(x-1)^3-(x+1)^2[(x-1)^3]'}{(x-1)^6}=\frac{(2x+2)(x-1)^3-3(x-1)^2(x+1)^2}{(x-1)^6}\)

\(=\frac{(2x+2)(x-1)-3(x+1)^2}{(x-1)^4}=\frac{-(x+1)(x+5)}{(x-1)^4}\)

d) 

\(y=1+(1-2x)^{\frac{3}{2}}\)

\(y'=\frac{3}{2}(1-2x)'(1-2x)^{\frac{3}{2}-1}=\frac{3}{2}(-2)(1-2x)^{\frac{1}{2}}=-3\sqrt{1-2x}\)

e) 

$y=(x-2)(x^2+3)^{\frac{1}{2}}=(x-2)'(x^2+3)^{\frac{1}{2}}+(x-2)\frac{1}{2}.(x^2+3)'(x^2+3)^{\frac{1}{2}-1}$

$=\sqrt{x^2+3}+\frac{x(x-2)}{\sqrt{x^2+3}}=\frac{2x^2-2x+3}{\sqrt{x^2+3}}$

f)

\(y=\frac{(4x+1)'\sqrt{x^2+2}-(4x+1)\sqrt{x^2+2}'}{x^2+2}=\frac{4\sqrt{x^2+2}-\frac{x(4x+1)}{\sqrt{x^2+2}}}{x^2+2}\)

\(=\frac{8-x}{\sqrt{(x^2+2)^3}}\)

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

b/ \(y'=5\left(\sqrt{x^3+1}-x^2+2\right)^4\left(\sqrt{x^3+1}-x^2+2\right)'=5\left(\sqrt{x^3+1}-x^2+2\right)^4\left(\dfrac{3x^2}{\sqrt{x^3+1}}-2x\right)\)c/ 

\(y'=14\left(x^6+2x-3\right)^6\left(x^6+2x-3\right)'=14\left(x^6+2x-3\right)^6\left(6x^5+2\right)\)

d/ \(y=\left(x^3-1\right)^{-\dfrac{5}{2}}\Rightarrow y'=-\dfrac{5}{2}\left(x^3-1\right)^{-\dfrac{7}{2}}\left(x^3-1\right)'=-\dfrac{15x^2}{2\sqrt{\left(x^3-1\right)^7}}\)

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

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

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

4/ \(y'=f'\left(x\right)=2x-\dfrac{2x}{x^4}=2x-\dfrac{2}{x^3}\)

\(y'=0\Leftrightarrow\dfrac{2x^4-2}{x^3}=0\Leftrightarrow x=\pm1\)

5/ \(y'=\dfrac{\dfrac{1}{2\sqrt{1+x}}}{2\sqrt{1+\sqrt{1+x}}}\Rightarrow f\left(x\right).f'\left(x\right)=\sqrt{1+\sqrt{1+x}}.\dfrac{1}{4\sqrt{1+x}.\sqrt{1+\sqrt{1+x}}}=\dfrac{1}{4\sqrt{1+x}}=\dfrac{1}{2\sqrt{2}}\)

\(\Leftrightarrow2\sqrt{1+x}=\sqrt{2}\Leftrightarrow1+x=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\)

Hãy nhớ câu tính đạo hàm này, bởi nó liên quan đến nguyên hàm sau này sẽ học

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

Gọi tiếp tuyến d qua A có dạng: \(y=k\left(x-m\right)+1\)

d là tiếp tuyến của (C) khi hệ sau có nghiệm: \(\left\{{}\begin{matrix}\dfrac{2-x}{x-1}=k\left(x-m\right)+1\\\dfrac{-1}{\left(x-1\right)^2}=k\end{matrix}\right.\)

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

\(\Leftrightarrow-x^2+3x-2=m+x^2-3x+1\)

\(\Leftrightarrow-2x^2+6x-3=m\) (1)

Để từ A kẻ được đúng 1 tiếp tuyến \(\Rightarrow\) (1) có đúng 1 nghiệm thỏa mãn \(x\ne1\)

TH1: (1) có 1 nghiệm bằng 1 và 1 nghiệm khác 1 \(\Rightarrow m=1\)

TH2: đường thẳng \(y=m\) cắt \(y=-2x^2+6x-3\) tại đúng 1 điểm

\(\Rightarrow m=\dfrac{3}{2}\)

Điểm có hoành độ bằng tung độ \(\Rightarrow x=\sqrt{2x^2-4}\) (\(x\ge0\))

\(\Leftrightarrow x^2=2x^2-4\Rightarrow x=2\)

Tọa độ tiếp điểm: \(\left(2;2\right)\)

\(f'\left(x\right)=\dfrac{2x}{\sqrt{2x^2-4}}\Rightarrow f'\left(2\right)=2\)

Tiếp tuyến: \(y=2\left(x-2\right)+2\Leftrightarrow y=2x-2\)