What is the derivative of f(x) = (x)(6^(-2x))f(x)=(x)(62x)?

What is the derivative of f(x) = (x)(6^(-2x))f(x)=(x)(62x)?

2 Answers

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

Explanation:

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))

Jun 28, 2018

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

Explanation:

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

Use the product rule :

(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)