
d := 10^A * Exp[-1/2* (t- t0)^2 * (10^σ)^-2] * (f/r)^-γ​
In[81]:=
d:=(10^A*Exp[-(t-t0)^2/(2*(10^σ)^2)])/(f/r)^γ

D[d, A]/d
alternate interpretations
Assuming "A" is a variable
|
Use as
a unit
instead
In[82]:=
D[d,A]/d
Out[82]=
log(10)

simp := D[d, σ] / d
In[85]:=
simp:=D[(10^A*Exp[-(t-t0)^2/(2*(10^σ)^2)])/(f/r)^γ,σ]/((10^A*Exp[-(t-t0)^2/(2*(10^σ)^2)])/(f/r)^γ)

Simplify[simp]
In[86]:=
Simplify[simp]
Out[86]=
-σ
100
log(10)
2
(t-t0)

simp === (t - t0)^2 * (10^(σ))^-2 * log(10)
alternate interpretations
Assuming "log" is the natural logarithm
|
Use
the base 10 logarithm
instead
In[87]:=
simp===((t-t0)^2*Log[10])/(10^σ)^2
Out[87]=
False

FullSimplify[(10^σ)^-2*(t-t0)^2*log(10)]
alternate interpretations
full Wolfram|Alpha results
Assuming "FullSimplify" is referring to algebraic simplifications
|
Use as
referring to a computation
instead
Assuming "log" is the natural logarithm
|
Use
the base 10 logarithm
instead
In[91]:=
Simplify[((t-t0)^2*Log[10])/(10^σ)^2]
Out[91]=
-2σ
10
log(10)
2
(t-t0)

FullSimplify[simp] === FullSimplify[100^(-σ)*log(10)*(t-t0)^2]

No Wolfram Language translation found.

100^(-σ) === (10^(2))^(-σ)
full Wolfram|Alpha results
In[94]:=
100^(-σ)===(10^2)^(-σ)
Out[94]=
True

10^(-2*σ) ===100^-σ
full Wolfram|Alpha results
In[95]:=
10^(-2*σ)===100^(-σ)
Out[95]=
False

D[a^-x, x]
step-by-step solution
related computations
full Wolfram|Alpha results
In[28]:=
D[a^(-x),x]
Out[28]=
-
-x
a
log(a)

D[a^x,x]
step-by-step solution
related computations
full Wolfram|Alpha results
In[1]:=
D[a^x,x]
Out[1]=
x
a
log(a)

FullSimplify[D[d, γ]/d]
alternate interpretations
Assuming "FullSimplify" is referring to a computation
|
Use as
referring to algebraic simplifications
instead
In[27]:=
FullSimplify[-Log[f/r]]
Out[27]=
-log
f
r