What is the derivative of #f(x) = (x)(6^(-2x))#?

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Answer 1 · 2018-06-28T17:16:19

#f'(x)=6^{-2x}(1-2x\ln(6))#

Using product rule of differentiation

#f(x)=x6^{-2x}#

#\frac{d}{dx}f(x)=\frac{d}{dx}(xe^{-2x})#

#\frac{df(x)}{dx}=x\frac{d}{dx}6^{-2x}+6^{-2x}\frac{d}{dx}x#

#=x6^{-2x}\ln(6)\cdot (-2)+6^{-2x}\cdot 1#

#=6^{-2x}(-2x\ln(6)+1)#

#=6^{-2x}(1-2x\ln(6))#

Answer 2 · 2018-06-28T17:18:10

#f'(x) = 6^(-2x)+xln(1/36)6^(-2x)#

Given: #f(x) = (x)(6^(-2x))#

#(d(u*v))/dx = (du)/dxv+u (dv)/dx#

where, #u = x# and #v = 6^(-2x)#

#(du)/dx = 1#

For #(dv)/dx#, we must use logarithmic differentiation. Start with:

#v = 6^(-2x)#

take the natural logarithm of both sides:

#ln(v) = ln(6^(-2x))#

Use the property of logarithms #ln(a^c) = cln(a)#:

#ln(v) = (-2x)ln(6)#

Differentiate both sides:

#1/v(dv)/dx=-2ln(6)#

#1/v(dv)/dx = ln(1/36)#

Multiply both sides by #v#:

#(dv)/dx = ln(1/36)v#

We know that #v = 6^(-2x)#:

#(dv)/dx = ln(1/36)6^(-2x)#

Substituting these back into the product rule:

#(d(x(6^(-2x))))/dx = 6^(-2x)+xln(1/36)6^(-2x)#

#f'(x) = 6^(-2x)+xln(1/36)6^(-2x)#