Chương 5: ĐẠO HÀM - Hỏi đáp

Lời giải:

Sử dụng liên hợp:

\(\lim_{x\to 0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}=\lim_{x\to 0}\frac{(\sqrt{x+1}-1)+(\sqrt{x+4}-2)}{x}\)

\(=\lim_{x\to 0}\frac{\frac{x}{\sqrt{x+1}+1}+\frac{x}{\sqrt{x+4}+2}}{x}=\lim_{x\to 0}\frac{1}{\sqrt{x+1}+1}+\frac{1}{\sqrt{x+4}+2}\)

\(=\frac{1}{\sqrt{0+1}+1}+\frac{1}{\sqrt{0+4}+2}=\frac{3}{4}\)

\(\sqrt{x}+\dfrac{1}{\sqrt{x}}+\dfrac{x^{10}}{10}=U+V+T\)

\(\left\{{}\begin{matrix}U^2=x;\\V^2=\dfrac{1}{x}\\Y'=U'+V'+T'\end{matrix}\right.\) \(\begin{matrix}\left(1\right)\\\left(2\right)\\\left(3\right)\end{matrix}\)

\(\left(1\right)\Leftrightarrow U'=\dfrac{1}{2U}=\dfrac{1}{2\sqrt{x}}\)

(2) \(\Leftrightarrow V'=\dfrac{-1}{x^2.2V}=\dfrac{-1}{2x^2.\dfrac{1}{\sqrt{x}}}=\dfrac{-1}{2.\sqrt[3]{x^2}}\)

\(\left(3\right)\Leftrightarrow Y'=\dfrac{1}{2\sqrt{x}}-\dfrac{1}{2\sqrt[3]{x^2}}+x^9\)

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

\(\Rightarrow y'=-5\left(2x-1\right)\left(x^2-x+1\right)^{-6}=\dfrac{5-10x}{\left(x^2-x+1\right)^6}\)

b: \(y=x^2+x^{\dfrac{3}{2}}+1\Rightarrow y'=2x+\dfrac{3}{2}x^{\dfrac{1}{2}}=2x+\dfrac{3\sqrt{x}}{2}\)

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

Lời giải:

Có \(y=\sqrt{(2x+1)^{2017}}=(2x+1)^{\frac{2017}{2}}\)

\(\Rightarrow y'=\frac{2017}{2}(2x+1)'(2x+1)^{\frac{2017}{2}-1}\)

\(=\frac{2017}{2}.2(2x+1)^{\frac{2015}{2}}=2017\sqrt{(2x+1)^{2015}}\)

ghi sai toán olớp 6 nhé ai giúp mình với

Lời giải:

Tìm đạo hàm theo biến $y$, bạn chỉ cần coi $x$ là một tham số rồi sử dụng công thức như bình thường thôi.

\(f(y)=y.e^{xy}.\sin y\)

\(\Rightarrow f'(y)=(y.e^{xy})'\sin y+y.e^{xy}(\sin y)'\)

\(=[y'.e^{xy}+y(e^{xy})']\sin y+y.e^{xy}.\cos y\)

\(=(e^{xy}+yxe^{xy})\sin y+y.e^{xy}\cos y\)

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Tính đạo hàm cấp 2.

Theo biến $x$

\(f(x)=e^{xy}\sin y\)

\(\Rightarrow f'(x)=\sin y(e^{xy})'=\sin y.ye^{xy}\)

\(\Rightarrow f''(x)=(y\sin y.e^{xy})'=y\sin y(e^{xy})'=y^2\sin y.e^{xy}\)

Theo biến $y$

\(f(y)=e^{xy}.\sin y\)

\(\Rightarrow f'(y)=(e^{xy})'\sin y+(\sin y)'e^{xy}\)

\(=x.e^{xy}\sin y+\cos y.e^{xy}\)

\(\Rightarrow f''(y)=(xe^{xy}.\sin y+\cos y.e^{xy})'\)

\(=(x.e^{xy}\sin y)'+(\cos y.e^{xy})'\)

\(=(x.e^{xy})'\sin y+(\sin y)'.xe^{xy}+(\cos y)'e^{xy}+\cos y(e^{xy})'\)

\(=x^2e^{xy}.\sin y+\cos y.x.e^{xy}-\sin y.e^{xy}+x\cos y.e^{xy}\)

5,479316961*10^13