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

Related questions
See all questions in Differentiating Exponential Functions with Other Bases
Impact of this question
1881 views around the world
You can reuse this answer
Creative Commons License