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

Question

2060 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-03T21:36:04

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

$y = x^3 2^x$

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

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