$$$\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}$$$的导数
您的输入
求$$$\frac{d}{dt} \left(\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right)$$$。
解答
对 $$$c = 7$$$ 和 $$$f{\left(t \right)} = \frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}$$$ 应用常数倍法则 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dt} \left(\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right)\right)} = {\color{red}\left(7 \frac{d}{dt} \left(\frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right)\right)}$$函数$$$\frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}$$$是两个函数$$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$和$$$g{\left(t \right)} = 9 t^{4} + 4 t^{2} + 49$$$的复合$$$f{\left(g{\left(t \right)} \right)}$$$。
应用链式法则 $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$:
$$7 {\color{red}\left(\frac{d}{dt} \left(\frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right)\right)} = 7 {\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right) \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)\right)}$$应用幂次法则 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$,其中 $$$n = - \frac{1}{2}$$$:
$$7 {\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right)\right)} \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right) = 7 {\color{red}\left(- \frac{1}{2 u^{\frac{3}{2}}}\right)} \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)$$返回到原变量:
$$- \frac{7 \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)}{2 {\color{red}\left(u\right)}^{\frac{3}{2}}} = - \frac{7 \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)}{2 {\color{red}\left(9 t^{4} + 4 t^{2} + 49\right)}^{\frac{3}{2}}}$$和/差的导数等于导数的和/差:
$$- \frac{7 {\color{red}\left(\frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)\right)}}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 {\color{red}\left(\frac{d}{dt} \left(9 t^{4}\right) + \frac{d}{dt} \left(4 t^{2}\right) + \frac{d}{dt} \left(49\right)\right)}}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$对 $$$c = 4$$$ 和 $$$f{\left(t \right)} = t^{2}$$$ 应用常数倍法则 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$:
$$- \frac{7 \left({\color{red}\left(\frac{d}{dt} \left(4 t^{2}\right)\right)} + \frac{d}{dt} \left(49\right) + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left({\color{red}\left(4 \frac{d}{dt} \left(t^{2}\right)\right)} + \frac{d}{dt} \left(49\right) + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$应用幂次法则 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$,其中 $$$n = 2$$$:
$$- \frac{7 \left(4 {\color{red}\left(\frac{d}{dt} \left(t^{2}\right)\right)} + \frac{d}{dt} \left(49\right) + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left(4 {\color{red}\left(2 t\right)} + \frac{d}{dt} \left(49\right) + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$常数的导数是$$$0$$$:
$$- \frac{7 \left(8 t + {\color{red}\left(\frac{d}{dt} \left(49\right)\right)} + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left(8 t + {\color{red}\left(0\right)} + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$对 $$$c = 9$$$ 和 $$$f{\left(t \right)} = t^{4}$$$ 应用常数倍法则 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$:
$$- \frac{7 \left(8 t + {\color{red}\left(\frac{d}{dt} \left(9 t^{4}\right)\right)}\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left(8 t + {\color{red}\left(9 \frac{d}{dt} \left(t^{4}\right)\right)}\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$应用幂次法则 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$,其中 $$$n = 4$$$:
$$- \frac{7 \left(8 t + 9 {\color{red}\left(\frac{d}{dt} \left(t^{4}\right)\right)}\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left(8 t + 9 {\color{red}\left(4 t^{3}\right)}\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$化简:
$$- \frac{7 \left(36 t^{3} + 8 t\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{14 t \left(9 t^{2} + 2\right)}{\left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$因此,$$$\frac{d}{dt} \left(\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right) = - \frac{14 t \left(9 t^{2} + 2\right)}{\left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$$。
答案
$$$\frac{d}{dt} \left(\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right) = - \frac{14 t \left(9 t^{2} + 2\right)}{\left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$$A