Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ \(y'=4\left(2x-3\right)^3.\left(2x-3\right)'=8\left(2x-3\right)^3\)
b/ \(y'=5cos^43x.\left(cos3x\right)'=-15cos^43x.sin3x\)
c/ \(y'=\frac{\left[cos\left(1-2x^2\right)\right]'}{2\sqrt{cos\left(1-2x^2\right)}}=\frac{-sin\left(1-2x^2\right).\left(1-2x^2\right)'}{2\sqrt{cos\left(1-2x^2\right)}}=\frac{2x.sin\left(1-2x^2\right)}{\sqrt{cos\left(1-2x^2\right)}}\)
d/ \(y'=\frac{\left(\frac{x+1}{x-1}\right)'}{2\sqrt{\frac{x+1}{x-1}}}=\frac{\frac{-2}{\left(x-1\right)^2}}{2\sqrt{\frac{x+1}{x-1}}}=-\frac{1}{\left(x-1\right)^2\sqrt{\frac{x+1}{x-1}}}\)
e/ \(y'=4\left(1+sin^2x\right)^3\left(1+sin^2x\right)'=8.sinx.cosx\left(1+sin^2x\right)^3=4sin2x.\left(1+sin^2x\right)^3\)
a/ \(y'=tan4x+\left(x+1\right)\left(tan4x\right)'=tan4x+\frac{4\left(x+1\right)}{cos^24x}\)
b/ \(y'=2cos\left(sin\sqrt{x}\right)\left[cos\left(sin\sqrt{x}\right)\right]'=2cos\left(sin\sqrt{x}\right).\left(-sin\left(sin\sqrt{x}\right)\right)\left(sin\sqrt{x}\right)'\)
\(=-\frac{sin\left(2sin\sqrt{x}\right).cos\sqrt{x}}{2\sqrt{x}}\)
c/ \(y'=\frac{\left(sin2x\right)'}{cos^2\left(sin2x\right)}=\frac{2cos2x}{cos^2\left(sin2x\right)}\)