Chương 5: ĐẠO HÀM

\(a+b=2\sqrt{ab}\)

=>\(a-2\sqrt{ab}+b=0\)

=>\(\left(\sqrt{a}-\sqrt{b}\right)^2=0\)

=>a=b

\(ln\sqrt{a}+ln\sqrt{b}=ln\left(\sqrt{ab}\right)=ln\left(\sqrt{a\cdot a}\right)=lna\)

\(\dfrac{1}{4}\left(lna+lnb\right)=\dfrac{1}{4}\left(lnab\right)=\dfrac{1}{4}\left(lna^2\right)=\dfrac{1}{4}\cdot2\cdot lna=\dfrac{1}{2}\cdot lna\)

=>Loại C

\(ln\left(\sqrt{a}+\sqrt{b}\right)=ln\left(\sqrt{a}+\sqrt{a}\right)=ln\left(2\sqrt{a}\right)\)

=>Loại B

\(ln\left(\dfrac{\sqrt{a}+\sqrt{b}}{2}\right)=ln\left(\dfrac{\sqrt{a}+\sqrt{a}}{2}\right)=ln\sqrt{a}=\dfrac{1}{2}\cdot lna\)

=>Chọn A

\(log_{ab}a^2=3\)

=>\(\dfrac{1}{log_{a^2}ab}=\dfrac{1}{3}\)

=>\(\dfrac{1}{\dfrac{1}{2}\cdot log_aab}=\dfrac{1}{3}\)

=>\(\dfrac{2}{log_aa+log_ab}=\dfrac{1}{3}\)

=>\(\dfrac{2}{1+log_ab}=\dfrac{1}{3}\)

=>\(1+log_ab=6\)

=>\(log_ab=5\)

\(T=log_{ab}\left(\sqrt[3]{\dfrac{a}{b}}\right)\)

\(=\dfrac{1}{3}\cdot log_{ab}\left(\dfrac{a}{b}\right)\)

\(=\dfrac{1}{3}\cdot log_{ab}\left(\dfrac{a^2}{ab}\right)\)

\(=\dfrac{1}{3}\left(log_{ab}a^2-1\right)\)

\(=\dfrac{1}{3}\left(3-1\right)=\dfrac{1}{3}\cdot2=\dfrac{2}{3}\)

=>Chọn D

Câu 42:

\(log_{abc}\left(ab^2c^3\right)\)

\(=log_{abc}\left(abc\cdot bc^2\right)\)

\(=1+log_{abc}\left(bc^2\right)\)

\(=1+log_{abc}b+log_{abc}c^2\)

\(=1+\dfrac{1}{log_babc}+2\cdot log_{abc}c\)

\(=1+\dfrac{1}{log_ba+1+log_bc}+\dfrac{2}{log_cabc}\)

\(=1+\dfrac{1}{1+log_ba+log_bc}+\dfrac{2}{log_ca+log_cb+1}\)

\(=1+\dfrac{1}{1+\dfrac{1}{m}+\dfrac{log_ac}{log_ab}}+\dfrac{2}{\dfrac{1}{log_ac}+\dfrac{log_ab}{log_ac}+1}\)

\(=1+\dfrac{1}{\dfrac{m+1}{m}+\dfrac{n}{m}}+\dfrac{2}{\dfrac{1}{n}+\dfrac{m}{n}+1}\)

\(=1+\dfrac{m}{m+1+n}+\dfrac{2}{\dfrac{1+m+n}{n}}\)

\(=1+\dfrac{m}{m+n+1}+\dfrac{2n}{m+n+1}=\dfrac{m+2n+m+n+1}{m+n+1}\)

\(=\dfrac{2m+3n+1}{m+n+1}\)

=>Chọn A

Câu 40:

\(M=a^{2016log_{a^2}2017}\)

\(=a^{2016\cdot\dfrac{1}{2}\cdot log_a2017}\)

\(=a^{1008\cdot log_a2017}=a^{log_a2017^{1008}}=2017^{1008}\)

=>Chọn D

Câu 12:

\(T=log_{a^2}b^6+log_a\sqrt{b}\)

\(=\dfrac{1}{2}log_ab^6+log_a\sqrt{b}\)

\(=\left(\dfrac{1}{2}\cdot6\right)log_ab+\dfrac{1}{2}log_ab\)

\(=3log_ab+\dfrac{1}{2}log_ab\)

\(=3\cdot2+\dfrac{1}{2}\cdot2\)

\(=7\)

⇒ Chọn B 

Coi như tất cả các biểu thức cần tính đạo hàm đều xác định.

1.

\(y'=2sin\sqrt{4x+3}.\left(sin\sqrt{4x+3}\right)'=2sin\sqrt{4x+3}.cos\sqrt{4x+3}.\left(\sqrt{4x+3}\right)'\)

\(=sin\left(2\sqrt{4x+3}\right).\dfrac{4}{2\sqrt{4x+3}}=\dfrac{2sin\left(2\sqrt{4x+3}\right)}{\sqrt{4x+3}}\)

2.

\(y'=3x^3+\dfrac{17}{x\sqrt{x}}\)

3.

\(y'=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\left(\dfrac{sin4x}{cos\left(x^2+2\right)}\right)'\)

\(=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\dfrac{4cos4x.cos\left(x^2+2\right)+2x.sin4x.sin\left(x^2+2\right)}{cos^2\left(x^2+2\right)}\)

4.

\(y'=-\dfrac{\left(\sqrt{sin^2\left(6-x\right)+4x}\right)'}{sin^2\left(6-x\right)+4x}=-\dfrac{\left[sin^2\left(6-x\right)+4x\right]'}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)

\(=-\dfrac{2sin\left(6-x\right).\left[sin\left(6-x\right)\right]'+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}=-\dfrac{-2sin\left(6-x\right).cos\left(6-x\right)+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)

\(=\dfrac{sin\left(12-2x\right)-4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)

5.

\(y'=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).\left[sin\left(\dfrac{2x-1}{4-x}\right)\right]'\)

\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).cos\left(\dfrac{2x-1}{4-x}\right).\left(\dfrac{2x-1}{4-x}\right)'\)

\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+x.sin\left(\dfrac{4x-2}{4-x}\right).\dfrac{7}{\left(4-x\right)^2}\)

8.

\(y=tan^33x-\left(sin2x+cos3x\right)^5\)

\(\Rightarrow y'=3tan^23x.\left(tan3x\right)'-5\left(sin2x+cos3x\right)^4.\left(sin2x+cos3x\right)'\)

\(=\dfrac{9.tan^23x}{cos^23x}-5\left(sin2x+cos3x\right)^4.\left(2cos2x-3sin3x\right)\)

9.

\(y'=6cot^55x.\left(cot5x\right)'-4cos^33x.\left(cos3x\right)'+3cos3x\)

\(=-\dfrac{30.cot^55x}{sin^25x}+12cos^33x.sin3x+3cos3x\)

a.

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

b.

\(y'=\dfrac{\left(3x^2+6x\right)\left(-4x^2+x\right)-\left(x^3+3x^2+2\right)\left(-8x+1\right)}{\left(-4x^2+x\right)^2}=\dfrac{-4x^4+2x^3+3x^2+16x-2}{\left(4x^2-x\right)^2}\)

c.

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

\(=\dfrac{3\left(x^3+3x^2+3\right)^2\left(3x^2+6x\right)}{2\sqrt{\left(x^3+3x^2+3\right)^3}}=\dfrac{9}{2}\left(x^2+2x\right)\sqrt{x^3+3x^2+3}\)

d.

\(y=\left(x^3-5\right)\left(x^2-4x+4\right)\Rightarrow y'=3x^2\left(x^2-4x+4\right)+\left(x^3-5\right)\left(2x-4\right)\)

e.

\(y'=-2sin\left(2x+3\right)-3cot^2\left(x^2+2x\right).\left[cot\left(x^2+2x\right)\right]'\)

\(=-2sin\left(2x+3\right)-3cot^2\left(x^2+2x\right).\dfrac{-\left(2x+2\right)}{sin^2\left(x^2+2x\right)}\)

f.

\(y'=3sin^2\left(1-2x\right).\left[sin\left(1-2x\right)\right]'+\dfrac{5-6x^2}{cos^2\left(5x-2x^3\right)}\)

\(=-6sin^2\left(1-2x\right).cos\left(1-2x\right)+\dfrac{5-6x^2}{cos^2\left(5x-2x^3\right)}\)

g.

\(y'=cos\left(2x^2-3x\right)^3.\left[\left(2x^2-3x\right)^3\right]'-3tan^2\sqrt{2x+7}.\left[tan\sqrt{2x+7}\right]'\)

\(=3\left(2x^2-3x\right)^2\left(4x-3\right).cos\left(2x^2-3x\right)^3-3tan^2\sqrt{2x+7}.\dfrac{1}{\sqrt{2x+7}.cos^2\sqrt{2x+7}}\)

h.

\(y'=4cos^3\left(2x-\dfrac{\pi}{4}\right)\left[cos\left(2x-\dfrac{\pi}{4}\right)\right]'.tan\left(\dfrac{x+\pi}{3}\right)+cos^4\left(2x-\dfrac{\pi}{4}\right).\dfrac{1}{3cos^2\left(\dfrac{x+\pi}{3}\right)}\)

\(=-8cos^3\left(2x-\dfrac{\pi}{4}\right).sin\left(2x-\dfrac{\pi}{4}\right)+cos^4\left(2x-\dfrac{\pi}{4}\right).\dfrac{1}{3cos^2\left(\dfrac{x+\pi}{3}\right)}\)

\(3^x+3^{-x}=2^{2023}\)

=>\(3^x+\dfrac{1}{3^x}=2^{2023}\)

=>\(3^{2x}+1=3^x\cdot2^{2023}\)

\(A=\dfrac{3^{6x}+3^{3x}+1}{3^{2x}\left(3^{2x}+1\right)}\)

\(=\dfrac{3^{3x}\left(3^{3x}+1\right)+1}{3^{2x}\cdot3^x\cdot2^{2023}}\)

\(=\dfrac{3^{3x}+2}{2^{2023}}\)

\(f'\left(x\right)=\dfrac{1}{40}\cdot2x\cdot\left(-1\right)=-\dfrac{1}{20}x\)

Câu 1:

\(y'=\dfrac{\left(x^2-3x+5\right)'\left(x^2-1\right)-\left(x^2-3x+5\right)\left(x^2-1\right)'}{\left(x^2-1\right)^2}\)

\(=\dfrac{\left(2x-3\right)\left(x^2-1\right)-2x\left(x^2-3x+5\right)}{\left(x^2-1\right)^2}\)

\(=\dfrac{2x^3-2x-3x^2+3-2x^3+6x^2-10x}{\left(x^2-1\right)^2}\)

\(=\dfrac{3x^2-12x}{\left(x^2-1\right)^2}\)