Asked by Aux
d dx [(e^x + x)^x].
Can someone please show me a step by step of how to do this. Also i'm stuck at (e^x+1)((e^x+x)^x * ln(e^x+x)) ... am i on the right track or?
Can someone please show me a step by step of how to do this. Also i'm stuck at (e^x+1)((e^x+x)^x * ln(e^x+x)) ... am i on the right track or?
Answers
Answered by
Reiny
y = (e^x + x)^x
I would take ln of both sides
ln y = x ln(e^x + x)
now use the product rule
(dy/dx)/x = x(e^x + 1) + ln(e^x + 1)
dy/dx = y(x(e^x + 1) + ln(e^x + 1))
I don't know what kind of simplification you need , but you could sub back the original y value
dy/dx = ( (e^x + x)^x )*(x(e^x + 1) + ln(e^x + 1))
I would take ln of both sides
ln y = x ln(e^x + x)
now use the product rule
(dy/dx)/x = x(e^x + 1) + ln(e^x + 1)
dy/dx = y(x(e^x + 1) + ln(e^x + 1))
I don't know what kind of simplification you need , but you could sub back the original y value
dy/dx = ( (e^x + x)^x )*(x(e^x + 1) + ln(e^x + 1))
Answered by
Aux
Just checking it on calculator. For some reason the answer is wrong.
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