Mathematics
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current location:Derivative function > Derivative function calculation history > Answer
    There are 1 questions in this calculation: for each question, the 2 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ 0.5kxx + 0.125jllsin(x)sin(x) - pl + plcos(x)\ with\ respect\ to\ x:\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = 0.5kx^{2} + 0.125jl^{2}sin(x)sin(x) + lpcos(x) - lp\\&\color{blue}{The\ first\ derivative\ function:}\\&\frac{d\left( 0.5kx^{2} + 0.125jl^{2}sin(x)sin(x) + lpcos(x) - lp\right)}{dx}\\=&0.5k*2x + 0.125jl^{2}cos(x)sin(x) + 0.125jl^{2}sin(x)cos(x) - lpsin(x) + 0\\=&kx + 0.125jl^{2}sin(x)cos(x) + 0.125jl^{2}sin(x)cos(x) - lpsin(x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( kx + 0.125jl^{2}sin(x)cos(x) + 0.125jl^{2}sin(x)cos(x) - lpsin(x)\right)}{dx}\\=&k + 0.125jl^{2}cos(x)cos(x) + 0.125jl^{2}sin(x)*-sin(x) + 0.125jl^{2}cos(x)cos(x) + 0.125jl^{2}sin(x)*-sin(x) - lpcos(x)\\ \end{split}\end{equation} \]





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