Question

Find the fourth derivative of f(x)=3^x+ln(x)-108x. f(4)(x)=​

Answer 1

`f(x) = 3^x + \ln x -108 x\\\\f'(x) = 3^x \ln 3 + \frac 1x - 108\\\\f''(x) = 3^x \cdot 0+ \ln 3\cdot 3^x\cdot \ln (3)-x^(-1-1)-0= 3^x \cdot \ln^2 3 -x^(-2)\\ \\f'''(x) = 3^x \cdot 0 + \ln^2 (3) \cdot 3^x \cdot \ln (3) +2x^(-2-1) = 3^x \ln^3 (3)+2x^(-3)\\\\f''''(x) = 3^x \cdot 0 + \ln^3 (3) \cdot 3^x \ln(3) -3\cdot 2 x^(-3-1) = 3^x\ln^4 (3)-6x^(-4) \\\\\text{Hence the fourth derivative of f(x) is}~~ 3^x\ln^4 (3)-6x^(-4).\\\\`

`\textbf{Note:}\\\\(d)/(dx)(uv) = u (dv)/(dx) + v(du)/(dx)\\\\(d)/(dx) \ln (x) = \frac 1x\\\\(d)/(dx) a^x = a^x \ln a\\\\(d)/(dx) x^n = nx^(n-1)\\\\(d)/(dx) (\text{Constant}) = 0`