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differentiate 3^x-1

Hi
i need help in this question

How we differentiate y=3^x-1

I have done like that

1/y dy /dx = ln(3^x-1)

1/y dy/dx = (x-1)ln (3)
dy/dx = y (x-1) ln (3)
dy/dx = (3^x-1)(x-1)ln(3)


BUT MARK SCHEME
shows 3^x-1 (ln(3))


where I get wrong?
Reply 1
Avatar for Qer
Qer
OP
16
@gcsemusicsucks
Original post by Qer
Hi
i need help in this question

How we differentiate y=3^x-1

y=3x−1y=3^{x-1} or y=3x−1y=3^x-1?



I have done like that

1/y dy /dx = \ln (3^x-1)
1/y dy/dx = (x-1) \ln (3)


If

ln⁡y=(x−1)ln⁡3 \ln y = (x-1) \ln3

then

1ydydx=ln⁡3\dfrac{1}{y} \dfrac{dy}{dx}= \ln 3

since (x−1)ln⁡3=xln⁡3−ln⁡3(x-1) \ln 3=x \ln 3- \ln 3 and the derivative of a constant is 0.
(edited 6 years ago)
Reply 3
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Qer
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16
Original post by BobbJo
y=3x−1y=3^{x-1} or y=3x−1y=3^x-1?


If

ln⁡y=(x−1)ln⁡3 \ln y = (x-1) \ln3

then

1ydydx=ln⁡3\dfrac{1}{y} \dfrac{dy}{dx}= \ln 3

since (x−1)ln⁡3=xln⁡3−ln⁡3(x-1) \ln 3=x \ln 3- \ln 3 and the derivative of a constant is 0.


y=3^{x-1}
(edited 6 years ago)
Reply 4
Avatar for Jarred
Jarred
18
Original post by Qer


1/y dy /dx = ln(3^x-1)


It looks like you've not differentiated the RHS.
(edited 6 years ago)
Alternative to implicit differentiation you could write 3x−1=e(x−1)ln(3)3^{x-1}=e^{(x-1)ln(3)} and use chain rule
Reply 6
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Qer
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16
Original post by Jarred
It looks like you've not differentiated the RHS.


yeah

you are right

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