Question

Find df/dx of given function f = f1.f2.f3


f1 = logex^4 # log with the base e

f2 = e^(2x-3)

f3 = x^(1/3) # cube root of x

Answer 1

$\therefore \: f = f1 \times f2 \times f3$

$\implies \: f = log_{e}( {x}^{4} ) \times {e}^{2x - 3} \times {x}^{ \frac{1}{3} }$

$\implies \: f = 4 log_{e}(x) \times {e}^{2x - 3} \times {x}^{ \frac{1}{3} }$

Taking log on both sides:

$\implies \: f = log \bigg \{4 log_{e}(x) \times {e}^{2x - 3} \times {x}^{ \frac{1}{3} } \bigg \}$

$\implies \: log(f)= 4 log \{ log_{e}(x) \} + log( {e}^{2x - 3}) + log({x}^{ \frac{1}{3} })$

$\implies \: log(f)= 4 log \{ log_{e}(x) \} +( 2x - 3) + \dfrac{1}{3} log(x)$

$\implies \: \dfrac{log(f)}{dx}= 4 \dfrac{d \: log \{ log_{e}(x) \}}{dx} + \dfrac{d \:(2x - 3)}{dx} + \dfrac{1}{3} \dfrac{d \: log(x)}{dx}$

$\implies \: \dfrac{1}{f} \times \dfrac{df}{dx} = \dfrac{4}{x log(x) } + 2 + \dfrac{1}{3x}$

$\implies \: \dfrac{df}{dx} = \bigg \{ \dfrac{4}{x log(x) } + 2 + \dfrac{1}{3x} \bigg \}f$

$\implies \: \dfrac{df}{dx} = \bigg \{ \dfrac{4}{x log(x) } + 2 + \dfrac{1}{3x} \bigg \} \bigg \{log_{e}( {x}^{4} ) \times {e}^{2x - 3} \times {x}^{ \frac{1}{3} } \bigg \}$

So, final answer is:

$\boxed{ \bf\: \dfrac{df}{dx} = \bigg \{ \dfrac{4}{x log(x) } + 2 + \dfrac{1}{3x} \bigg \} \bigg \{log_{e}( {x}^{4} ) \times {e}^{2x - 3} \times {x}^{ \frac{1}{3} } \bigg \}}$