How do you differentiate #y = x^3*2^x#?

1 Answer
Steve M
Jan 3, 2017

# dy/dx= 3x^2 2^x + x^3 2^xln(2)#

Explanation:

#y = x^3 2^x #

Take Natural logs:

# ln y = ln(x^3 2^x)#
# :. ln y = ln(x^3) + ln(2^x)#
# :. ln y = 3ln(x) + xln(2)#

Differentiate Implicitly:

# 1/y dy/dx= 3/x + ln(2)#

# :. 1/(x^3 2^x) dy/dx= 3/x + ln(2)#

# :. dy/dx= (x^3 2^x){ 3/x + ln(2) }#
# :. dy/dx= 3x^2 2^x + x^3 2^xln(2)#

You could also use the product rule

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