Bộ 10 Đề kiểm tra học kì I Chuyên đề toán 11: Kiểm tra 15 phút có đáp án (Đề 9)

-4x-3

-4x+3

4x+3

4x-3

$y^{\text{'}}=5{\left(1-x^3\right)}^4$

$y=-15x^2{\left(1-x^3\right)}^4$

$y^{\text{'}}=-3{\left(1-x^3\right)}^4$

$y^{\text{'}}=-5x^2{\left(1-x^3\right)}^4$

$y^{\text{'}}=\cos x$

$y^{\text{'}}=-\cos x$

$y^{\text{'}}=-\sin x$

$y^{\text{'}}=\frac{1}{\cos x}$

$y^{\text{'}}=\sin x\left(3{\cos }^{2}x-1\right)$

$y\text{'}=\sin x\left(3{\cos }^{2}x+1\right)$

$y\text{'}=\sin x\left({\cos }^{2}x+1\right)$

$y\text{'}=\sin x\left({\cos }^{2}x-1\right)$

$y^{\text{'}}=\frac{1+{\cot }^{2}2x}{\sqrt{\cot 2x}}$

$y^{\text{'}}=\frac{-\left(1+{\cot }^{2}2x\right)}{\sqrt{\cot 2x}}$

$y^{\text{'}}=\frac{1+{\tan }^{2}2x}{\sqrt{\cot 2x}}$

$y^{\text{'}}=\frac{-\left(1+{\tan }^{2}2x\right)}{\sqrt{\cot 2x}}$

$\left\{-1;2\right\}$

$\left\{-1;3\right\}$

$\left\{0;4\right\}$

$\left\{1;2\right\}$

$dy=\left(x\cos x-\sin x\right)dx$

$dy=\left(x\cos x\right)dx$

$dy=\left(\cos x-\sin x\right)dx$

$dy=\left(x\sin x\right)dx$

$y^{\text{'}\text{'}}=\frac{1}{\left(2x+5\right)\sqrt{2x+5}}$

$y^{\text{'}\text{'}}=\frac{1}{\sqrt{2x+5}}$

$y^{\text{'}\text{'}}=-\frac{1}{\left(2x+5\right)\sqrt{2x+5}}$

$y"=-\frac{1}{\sqrt{2x+5}}$

$y=-3x+8$

$y=-3x+6$

$y=3x-8$

$y=3x-6$

$y=-3x-3;y=-3x-11$

$y=-3x-3;y=-3x+11$

$y=-3x+3;y=-3x-11$

$y=-3x-3;y=3x-11$