Đạo hàm của hàm số \(y=\log_{\sqrt{2}}\left(x.3^{2x}+1\right)\) là
\(y'=\dfrac{\left(x\ln81+2\right).3^{2x}}{\left(x.3^{2x}+1\right)\ln2}\).\(y'=\dfrac{3^{2x}\ln9+1}{\left(x.3^{2x}+1\right)\ln\sqrt{2}}\).\(y'=\dfrac{\left(x.\ln3+1\right)3^{2x}}{\left(x.3^{2x}+1\right)\ln\sqrt{2}}\).\(y'=\dfrac{3^{2x}+4x^2.3^{2x-1}}{\left(x.3^{2x}+1\right)\ln\sqrt{2}}\).Hướng dẫn giải:Áp dụng công thức \(\left(\log_au\right)'=\dfrac{u'}{u.\ln a}\) ta có:
\(y'=\dfrac{\left(x.3^{2x}+1\right)'}{\left(x.3^{2x}+1\right).\ln\sqrt{2}}\)
\(=\dfrac{x.2.3^{2x}\ln3+3^{2x}}{\left(x.3^{2x}+1\right).\ln\sqrt{2}}\)
\(=\dfrac{\left(2x\ln3+1\right)3^{2x}}{\left(x.3^{2x}+1\right).\ln2^{\frac{1}{2}}}\)
\(=\dfrac{\left(2x\ln3+1\right)3^{2x}}{\frac{1}{2}\left(x.3^{2x}+1\right).\ln2}\)
\(=\dfrac{2\left(2x\ln3+1\right)3^{2x}}{\left(x.3^{2x}+1\right).\ln2}\)
\(=\dfrac{\left(4x\ln3+2\right)3^{2x}}{\left(x.3^{2x}+1\right).\ln2}\)
\(=\dfrac{\left(x\ln3^4+2\right)3^{2x}}{\left(x.3^{2x}+1\right).\ln2}\)
\(=\dfrac{\left(x\ln81+2\right)3^{2x}}{\left(x.3^{2x}+1\right).\ln2}\).
