Differentiation rules: differentiation of elementary functions d d d -{f(x) g(x)}= f(x) 8(x) + g(x). dx f(x),, assuming f and g are differentiable. dx 4.12. Prove the formula dx By definition, d f(x +Ax) g (x +Ax) – f(x) g(x) {f(x) g(x)}= lim dx Ax-0 Ax f(x+Ax) {g (x +Ax) – g(x)}+g(x) {f(x+Ax) – f(x)} lim Ax-0 Ax g(x +Ax) - g(x) f(x+ Ax) - f(x) = lim f(x +Ax) + lim g(x) Ax→0 Ax Ax→0 Ax d d = f(x) 8(x)+ g(x) f(x) dx dx Another method: Let u = f(x), v = g(x). Then Au = f(x + Ax) – f(x) and A v = g(x + Ax) – g(x); i.e., f(x + Ar) = u + Au, g(x + Ar) = v + Av. Thus, d uv = lim (и + Ди)(0 + дu) — w uAv +vAu + AuAv lim dx Ar→0 Ar Ar0 Ax dv du Δυ = lim u Ar Au Au Av = u- + Ax Ar-0 Ar dx dx where it is noted that Au →0 as Ar → 0, since v is supposed differentiable and thus continuous.
Differentiation rules: differentiation of elementary functions d d d -{f(x) g(x)}= f(x) 8(x) + g(x). dx f(x),, assuming f and g are differentiable. dx 4.12. Prove the formula dx By definition, d f(x +Ax) g (x +Ax) – f(x) g(x) {f(x) g(x)}= lim dx Ax-0 Ax f(x+Ax) {g (x +Ax) – g(x)}+g(x) {f(x+Ax) – f(x)} lim Ax-0 Ax g(x +Ax) - g(x) f(x+ Ax) - f(x) = lim f(x +Ax) + lim g(x) Ax→0 Ax Ax→0 Ax d d = f(x) 8(x)+ g(x) f(x) dx dx Another method: Let u = f(x), v = g(x). Then Au = f(x + Ax) – f(x) and A v = g(x + Ax) – g(x); i.e., f(x + Ar) = u + Au, g(x + Ar) = v + Av. Thus, d uv = lim (и + Ди)(0 + дu) — w uAv +vAu + AuAv lim dx Ar→0 Ar Ar0 Ax dv du Δυ = lim u Ar Au Au Av = u- + Ax Ar-0 Ar dx dx where it is noted that Au →0 as Ar → 0, since v is supposed differentiable and thus continuous.
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4.12) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text.
Transcribed Image Text:Differentiation rules: differentiation of elementary functions
d
Prove the formula
dx
By definition,
d
d
{f(x) g(x)}=
)}= f(x)g(x)+ g(x) f(x),, assuming fand g are differentiable.
4.12.
dx
dx
d
f(x +Ax) g (x +Ax) – f(x) g(x)
{f(x) g(x)}= lim
dx
Ax-0
Ax
f(x+Ax) {g (x +Ax) – g(x)}+g(x) {f(x+Ax) – f(x)}
= lim
Ax-0
Ax
g(x +Ax) - g(x)
f(x+ Ax) – f(x)
= lim f(x+Ax)
+ lim g(x)
Ax→0
Ax
Ax→0
Ax
d
d
= f(x)8(x)+ g(x)
f(x)
dx
dx
Another method:
Letu =f() , υ = g). Then Δ= f(x + Δι)-f) and Δυ- ( + Δ)-g(); ic., fr + Δε) =u+ Δι,
g( + Δ)=0 + Δυ. Thus ,
d
uv = lim
(μ+ Δυ) ( + Δυ)-ω
uAv +vAu + AuAv
lim
dx
Ar-0
Ar
Ar-0
Ax
do
du
+v
dx
Δυ
Au
+v=+
Ar
Δι
Av = u-
Ar
= lim u
Ar0
Ar
dx
where it is noted that Au → 0 as Ar → 0, since v is supposed differentiable and thus continuous.
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