How do you find the dy/dx of an equation like y= 2^(x-5)?
2006-11-02 22:39:21 · 7 answers · asked by rain-shadow 2 in Science & Mathematics ➔ Mathematics
taking log on both sides
lny=(x-5)ln2
1/y(dy/dx)=ln2
dy/dx=yln2=2^(x-5)ln2
2006-11-02 22:43:03 · answer #1 · answered by raj 7 · 0⤊ 0⤋
y=2^(x-5)
take log on both sides
log y=(x-5)log 2
differentiate wrt x
1/y dy/dx = log2
dy/dx=y log2
2006-11-02 23:27:33 · answer #2 · answered by . 3 · 0⤊ 0⤋
Take the natural log of both sides
Ln y = (x-5) Ln 2
Now take the derivative of both sides.
y' / y = Ln 2
y' = Ln 2 * y
y' = (Ln 2) * 2^(x-5)
2006-11-02 22:43:34 · answer #3 · answered by z_o_r_r_o 6 · 0⤊ 0⤋
Refer to the formula sheet and get the generalize formaula.
Apply subsitition method to do the rest of the workings
2006-11-03 02:32:24 · answer #4 · answered by Si J 2 · 0⤊ 0⤋
y = 2^(x - 5)
So ln y = (x - 5) ln2
So y = e^[(x - 5)ln2}
dy/dx = ln2 . e^[(x - 5)ln2}
= ln2 . 2^(x-5)
2006-11-02 22:44:51 · answer #5 · answered by Wal C 6 · 0⤊ 0⤋
rewrite the equation so
y(x)=2^(x-5) & a^z=e^(z*ln(a)) =>
e^[(x-5)ln(2)]=
y(u)=e^u & u=(x-5)*ln2
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use the chain rule dy/dx=(dy/du)*(du/dx)
dy/du= e^u=e^[(x-5)ln(2)]=2^(x-5)
du/dx=ln(2)
dy/dx=ln(2)2^(x-5)
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2006-11-02 22:59:14 · answer #6 · answered by Broden 4 · 1⤊ 0⤋
take ln (natural logirithm) both sides and then diffrentaite
2006-11-02 22:50:00 · answer #7 · answered by Charu Chandra Goel 5 · 0⤊ 0⤋