Turunan dari $$$\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}$$$
Kalkulator terkait: Kalkulator Diferensiasi Logaritmik, Kalkulator Diferensiasi Implisit dengan Langkah-langkah
Masukan Anda
Temukan $$$\frac{d}{dt} \left(\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right)$$$.
Solusi
Terapkan aturan kelipatan konstanta $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ dengan $$$c = 7$$$ dan $$$f{\left(t \right)} = \frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}$$$:
$${\color{red}\left(\frac{d}{dt} \left(\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right)\right)} = {\color{red}\left(7 \frac{d}{dt} \left(\frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right)\right)}$$Fungsi $$$\frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}$$$ merupakan komposisi $$$f{\left(g{\left(t \right)} \right)}$$$ dari dua fungsi $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$ dan $$$g{\left(t \right)} = 9 t^{4} + 4 t^{2} + 49$$$.
Terapkan aturan rantai $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$:
$$7 {\color{red}\left(\frac{d}{dt} \left(\frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right)\right)} = 7 {\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right) \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)\right)}$$Terapkan aturan pangkat $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ dengan $$$n = - \frac{1}{2}$$$:
$$7 {\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right)\right)} \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right) = 7 {\color{red}\left(- \frac{1}{2 u^{\frac{3}{2}}}\right)} \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)$$Kembalikan ke variabel semula:
$$- \frac{7 \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)}{2 {\color{red}\left(u\right)}^{\frac{3}{2}}} = - \frac{7 \frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)}{2 {\color{red}\left(9 t^{4} + 4 t^{2} + 49\right)}^{\frac{3}{2}}}$$Turunan dari jumlah/selisih adalah jumlah/selisih dari turunan:
$$- \frac{7 {\color{red}\left(\frac{d}{dt} \left(9 t^{4} + 4 t^{2} + 49\right)\right)}}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 {\color{red}\left(\frac{d}{dt} \left(9 t^{4}\right) + \frac{d}{dt} \left(4 t^{2}\right) + \frac{d}{dt} \left(49\right)\right)}}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$Terapkan aturan kelipatan konstanta $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ dengan $$$c = 4$$$ dan $$$f{\left(t \right)} = t^{2}$$$:
$$- \frac{7 \left({\color{red}\left(\frac{d}{dt} \left(4 t^{2}\right)\right)} + \frac{d}{dt} \left(49\right) + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left({\color{red}\left(4 \frac{d}{dt} \left(t^{2}\right)\right)} + \frac{d}{dt} \left(49\right) + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$Terapkan aturan pangkat $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ dengan $$$n = 2$$$:
$$- \frac{7 \left(4 {\color{red}\left(\frac{d}{dt} \left(t^{2}\right)\right)} + \frac{d}{dt} \left(49\right) + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left(4 {\color{red}\left(2 t\right)} + \frac{d}{dt} \left(49\right) + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$Turunan dari suatu konstanta adalah $$$0$$$:
$$- \frac{7 \left(8 t + {\color{red}\left(\frac{d}{dt} \left(49\right)\right)} + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left(8 t + {\color{red}\left(0\right)} + \frac{d}{dt} \left(9 t^{4}\right)\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$Terapkan aturan kelipatan konstanta $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ dengan $$$c = 9$$$ dan $$$f{\left(t \right)} = t^{4}$$$:
$$- \frac{7 \left(8 t + {\color{red}\left(\frac{d}{dt} \left(9 t^{4}\right)\right)}\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left(8 t + {\color{red}\left(9 \frac{d}{dt} \left(t^{4}\right)\right)}\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$Terapkan aturan pangkat $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ dengan $$$n = 4$$$:
$$- \frac{7 \left(8 t + 9 {\color{red}\left(\frac{d}{dt} \left(t^{4}\right)\right)}\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{7 \left(8 t + 9 {\color{red}\left(4 t^{3}\right)}\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$Sederhanakan:
$$- \frac{7 \left(36 t^{3} + 8 t\right)}{2 \left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}} = - \frac{14 t \left(9 t^{2} + 2\right)}{\left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$Dengan demikian, $$$\frac{d}{dt} \left(\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right) = - \frac{14 t \left(9 t^{2} + 2\right)}{\left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$$.
Jawaban
$$$\frac{d}{dt} \left(\frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right) = - \frac{14 t \left(9 t^{2} + 2\right)}{\left(9 t^{4} + 4 t^{2} + 49\right)^{\frac{3}{2}}}$$$A