Mathematics
语言:中文
Language:English

current location:Derivative function > Derivative function calculation history > Answer
    There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ lg(10 - Aarctan(ax))\ with\ respect\ to\ x:\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = lg(-Aarctan(ax) + 10)\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( lg(-Aarctan(ax) + 10)\right)}{dx}\\=&\frac{(-A(\frac{(a)}{(1 + (ax)^{2})}) + 0)}{ln{10}(-Aarctan(ax) + 10)}\\=&\frac{-Aa}{(-Aarctan(ax) + 10)(a^{2}x^{2} + 1)ln{10}}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-Aa}{(-Aarctan(ax) + 10)(a^{2}x^{2} + 1)ln{10}}\right)}{dx}\\=&\frac{-(\frac{-(-A(\frac{(a)}{(1 + (ax)^{2})}) + 0)}{(-Aarctan(ax) + 10)^{2}})Aa}{(a^{2}x^{2} + 1)ln{10}} - \frac{(\frac{-(a^{2}*2x + 0)}{(a^{2}x^{2} + 1)^{2}})Aa}{(-Aarctan(ax) + 10)ln{10}} - \frac{Aa*-0}{(-Aarctan(ax) + 10)(a^{2}x^{2} + 1)ln^{2}{10}}\\=&\frac{-A^{2}a^{2}}{(-Aarctan(ax) + 10)^{2}(a^{2}x^{2} + 1)^{2}ln{10}} + \frac{2Aa^{3}x}{(-Aarctan(ax) + 10)(a^{2}x^{2} + 1)^{2}ln{10}}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{-A^{2}a^{2}}{(-Aarctan(ax) + 10)^{2}(a^{2}x^{2} + 1)^{2}ln{10}} + \frac{2Aa^{3}x}{(-Aarctan(ax) + 10)(a^{2}x^{2} + 1)^{2}ln{10}}\right)}{dx}\\=&\frac{-(\frac{-2(-A(\frac{(a)}{(1 + (ax)^{2})}) + 0)}{(-Aarctan(ax) + 10)^{3}})A^{2}a^{2}}{(a^{2}x^{2} + 1)^{2}ln{10}} - \frac{(\frac{-2(a^{2}*2x + 0)}{(a^{2}x^{2} + 1)^{3}})A^{2}a^{2}}{(-Aarctan(ax) + 10)^{2}ln{10}} - \frac{A^{2}a^{2}*-0}{(-Aarctan(ax) + 10)^{2}(a^{2}x^{2} + 1)^{2}ln^{2}{10}} + \frac{2(\frac{-(-A(\frac{(a)}{(1 + (ax)^{2})}) + 0)}{(-Aarctan(ax) + 10)^{2}})Aa^{3}x}{(a^{2}x^{2} + 1)^{2}ln{10}} + \frac{2(\frac{-2(a^{2}*2x + 0)}{(a^{2}x^{2} + 1)^{3}})Aa^{3}x}{(-Aarctan(ax) + 10)ln{10}} + \frac{2Aa^{3}}{(-Aarctan(ax) + 10)(a^{2}x^{2} + 1)^{2}ln{10}} + \frac{2Aa^{3}x*-0}{(-Aarctan(ax) + 10)(a^{2}x^{2} + 1)^{2}ln^{2}{10}}\\=&\frac{-2A^{3}a^{3}}{(-Aarctan(ax) + 10)^{3}(a^{2}x^{2} + 1)^{3}ln{10}} + \frac{6A^{2}a^{4}x}{(-Aarctan(ax) + 10)^{2}(a^{2}x^{2} + 1)^{3}ln{10}} - \frac{8Aa^{5}x^{2}}{(a^{2}x^{2} + 1)^{3}(-Aarctan(ax) + 10)ln{10}} + \frac{2Aa^{3}}{(-Aarctan(ax) + 10)(a^{2}x^{2} + 1)^{2}ln{10}}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{-2A^{3}a^{3}}{(-Aarctan(ax) + 10)^{3}(a^{2}x^{2} + 1)^{3}ln{10}} + \frac{6A^{2}a^{4}x}{(-Aarctan(ax) + 10)^{2}(a^{2}x^{2} + 1)^{3}ln{10}} - \frac{8Aa^{5}x^{2}}{(a^{2}x^{2} + 1)^{3}(-Aarctan(ax) + 10)ln{10}} + \frac{2Aa^{3}}{(-Aarctan(ax) + 10)(a^{2}x^{2} + 1)^{2}ln{10}}\right)}{dx}\\=&\frac{-2(\frac{-3(-A(\frac{(a)}{(1 + (ax)^{2})}) + 0)}{(-Aarctan(ax) + 10)^{4}})A^{3}a^{3}}{(a^{2}x^{2} + 1)^{3}ln{10}} - \frac{2(\frac{-3(a^{2}*2x + 0)}{(a^{2}x^{2} + 1)^{4}})A^{3}a^{3}}{(-Aarctan(ax) + 10)^{3}ln{10}} - \frac{2A^{3}a^{3}*-0}{(-Aarctan(ax) + 10)^{3}(a^{2}x^{2} + 1)^{3}ln^{2}{10}} + \frac{6(\frac{-2(-A(\frac{(a)}{(1 + (ax)^{2})}) + 0)}{(-Aarctan(ax) + 10)^{3}})A^{2}a^{4}x}{(a^{2}x^{2} + 1)^{3}ln{10}} + \frac{6(\frac{-3(a^{2}*2x + 0)}{(a^{2}x^{2} + 1)^{4}})A^{2}a^{4}x}{(-Aarctan(ax) + 10)^{2}ln{10}} + \frac{6A^{2}a^{4}}{(-Aarctan(ax) + 10)^{2}(a^{2}x^{2} + 1)^{3}ln{10}} + \frac{6A^{2}a^{4}x*-0}{(-Aarctan(ax) + 10)^{2}(a^{2}x^{2} + 1)^{3}ln^{2}{10}} - \frac{8(\frac{-3(a^{2}*2x + 0)}{(a^{2}x^{2} + 1)^{4}})Aa^{5}x^{2}}{(-Aarctan(ax) + 10)ln{10}} - \frac{8(\frac{-(-A(\frac{(a)}{(1 + (ax)^{2})}) + 0)}{(-Aarctan(ax) + 10)^{2}})Aa^{5}x^{2}}{(a^{2}x^{2} + 1)^{3}ln{10}} - \frac{8Aa^{5}*2x}{(a^{2}x^{2} + 1)^{3}(-Aarctan(ax) + 10)ln{10}} - \frac{8Aa^{5}x^{2}*-0}{(a^{2}x^{2} + 1)^{3}(-Aarctan(ax) + 10)ln^{2}{10}} + \frac{2(\frac{-(-A(\frac{(a)}{(1 + (ax)^{2})}) + 0)}{(-Aarctan(ax) + 10)^{2}})Aa^{3}}{(a^{2}x^{2} + 1)^{2}ln{10}} + \frac{2(\frac{-2(a^{2}*2x + 0)}{(a^{2}x^{2} + 1)^{3}})Aa^{3}}{(-Aarctan(ax) + 10)ln{10}} + \frac{2Aa^{3}*-0}{(-Aarctan(ax) + 10)(a^{2}x^{2} + 1)^{2}ln^{2}{10}}\\=&\frac{-6A^{4}a^{4}}{(-Aarctan(ax) + 10)^{4}(a^{2}x^{2} + 1)^{4}ln{10}} + \frac{24A^{3}a^{5}x}{(-Aarctan(ax) + 10)^{3}(a^{2}x^{2} + 1)^{4}ln{10}} - \frac{36A^{2}a^{6}x^{2}}{(-Aarctan(ax) + 10)^{2}(a^{2}x^{2} + 1)^{4}ln{10}} + \frac{8A^{2}a^{4}}{(-Aarctan(ax) + 10)^{2}(a^{2}x^{2} + 1)^{3}ln{10}} + \frac{48Aa^{7}x^{3}}{(a^{2}x^{2} + 1)^{4}(-Aarctan(ax) + 10)ln{10}} - \frac{8A^{2}a^{6}x^{2}}{(a^{2}x^{2} + 1)^{4}(-Aarctan(ax) + 10)^{2}ln{10}} - \frac{16Aa^{5}x}{(-Aarctan(ax) + 10)(a^{2}x^{2} + 1)^{3}ln{10}} - \frac{8Aa^{5}x}{(a^{2}x^{2} + 1)^{3}(-Aarctan(ax) + 10)ln{10}}\\ \end{split}\end{equation} \]





Your problem has not been solved here? Please take a look at the  hot problems !


Return