derivative find of f(x) = lnx/e^×2
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0
Answer:
Thus, the derivative of ln x2 is 2/x. Note this result agrees with the plots of tangent lines for both positive and negative x. For x = 2, the derivative is 2/2 = 1, which agrees with the plot.
Answered by
0
Answer:
Weget
f(x)=ln(lnx)+
(lnx)
2
1
∫f(x)dx=∫(ln(lnx)+
(lnx)
2
1
)dx
x=e
t
,dx=e
t
dt
=∫(lnt+
t
2
1
)e
t
dt
=∫e
t
(lnt+
t
1
)dt+∫e
t
(
t
2
1
−
t
1
)dt
=∫e
t
(lnt+
dt
d(lnt)
)dt+∫e
t
(−
t
1
+
dt
d
(
t
−1
)dt)
=e
t
lnt−
t
e
t
+C
∫fdx=xln(lnx)−
lnx
x
+C
y=xln(lnx)
1
−e
+cput(e,e)
e=eln(1)−
1
e
+C
c=2e
y=xln(lnx)
lnx
−x
+2e
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