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

Answer 1

Ezra Adams

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

#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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Answer 2

Evelyn Agard

To differentiate ( y = x^3 \times 2^x ), you can use the product rule and the chain rule. The derivative will be:

[ y' = 3x^2 \times 2^x + x^3 \times \ln(2) \times 2^x ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.