Pembuktian dan Penurunan Rumus Turunan

martha yunanda turunan

Sebagaimana telah dijelaskan bahwasanya turunan ini didapat dari pendekatan limit. Pendekatan limit yang dimaksud adalah

$f^\prime (x) = \displaystyle \lim_{ h \to 0 } \frac{f(x+ h ) - f(x)}{h} \\ \text {dengan syarat nilai limit ada}$

#Pembuktian Rumus Turunan y=f(x)=k
Untuk penurunan rumus turunan y = f(x)= k , berikut uraiannya.

$y = k \rightarrow f(x) = k$

$f(x+h) = k$

$\begin{align} f^\prime (x) & = \displaystyle \lim_{ h \to 0 } \frac{f(x+ h ) - f(x)}{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ k - k}{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ 0}{h} \\ & = \displaystyle \lim_{ h \to 0 } 0 \\ & = 0 \end{align}$
Terbukti :

$f(x) = k \rightarrow f^\prime (x) = 0$

#Pembuktian rumus Turunan

$y = ax^n \rightarrow y^\prime = n.a.x^{n-1}$
Beberapa rumus lain yang digunakan dalam pembuktian ini,
Binomial Newton

$(x + h)^n = x^n + C_1^n x^{n-1}h^1 + C_2^n x^{n-2}h^2 + …+ C_{n-1}^n x^{n}h^{n-1} + h^n$
Rumus Kombinasi :

$C_r^n = \frac{n!}{(n-r)!r!}$
Sehingga :

$C_1^n = \frac{n!}{(n-1)!.1!} = \frac{n.n(n-1)!}{(n-1)!} = n$
Dengan

$n! = n.(n-1).(n-2)….3.2.1 \,$ . Misalkan :

$5! = 5.4.3.2.1 = 120$
Uraian Pembuktian:

$y = ax^n \rightarrow f(x) = ax^n$

$f(x+h) = a(x+h)^n = a(x^n + C_1^n x^{n-1}h^1 + C_2^n x^{n-2}h^2 + …+ C_{n-1}^n x^{n}h^{n-1} + h^n)$

$f(x+h) = ax^n + aC_1^n x^{n-1}h^1 + aC_2^n x^{n-2}h^2 + …+ aC_{n-1}^n x^{n}h^{n-1} + ah^n$
diturunkan:

$\begin{align} f^\prime (x) & = \displaystyle \lim_{ h \to 0 } \frac{f(x+ h ) - f(x)}{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ (ax^n + aC_1^n x^{n-1}h^1 + aC_2^n x^{n-2}h^2 + …+ aC_{n-1}^n x^{n}h^{n-1} + ah^n) - ax^n}{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ aC_1^n x^{n-1}h^1 + aC_2^n x^{n-2}h^2 + …+ aC_{n-1}^n x^{n}h^{n-1} + ah^n }{h} \\ & = \displaystyle \lim_{ h \to 0 } aC_1^n x^{n-1} + aC_2^n x^{n-2}h^1 + …+ aC_{n-1}^n x^{n}h^{n-2} + ah^{n-1} \\ & = aC_1^n x^{n-1} + aC_2^n x^{n-2}.0 + …+ aC_{n-1}^n x^{n}.0^{n-2} + a.0^{n-1} \\ & = aC_1^n x^{n-1} + 0 + …+ 0 + 0 \\ & = aC_1^n x^{n-1} \\ & = an x^{n-1} \end{align}$
Terbukti :

$f(x) = ax^n \rightarrow f^\prime (x) = n.a.x^{n-1} = nax^{n-1}$
Pembuktian rumus Turunan :

$y = U \pm V \rightarrow y^\prime = U^\prime \pm V^\prime$

$f(x) = U(x) + V(x) \rightarrow f^\prime (x) = U^\prime (x) + V^\prime (x)$
Fungsinya:

$f(x) = U(x) + V(x)$

$f(x+h) = U(x+h) + V(x+h)$
Diturunkan :

$\begin{align} f^\prime (x) & = \displaystyle \lim_{ h \to 0 } \frac{f(x+ h ) - f(x)}{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ [U(x+h) + V(x+h)] - [U(hx) + V(x)] }{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ U(x+h) - U(x) + V(x+h) - V(x) }{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ U(x+h) - U(x) }{h} + \frac{ V(x+h) - V(x) }{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ U(x+h) - U(x) }{h} + \displaystyle \lim_{ h \to 0 } \frac{ V(x+h) - V(x) }{h} \\ & = U^\prime (x) + V^\prime (x) \end{align}$
Terbukti :

$f(x) = U(x) + V(x) \rightarrow f^\prime (x) = U^\prime (x) + V^\prime (x)$

#Pembuktian rumus Turunan : U.V= U’V+UV’
Fungsinya :

$y = U.V \rightarrow f(x) = U(x).V(x)$

$f(x+h) = U(x+h).V(x+h)$
Diturunkan :

$\begin{align} f^\prime (x) & = \displaystyle \lim_{ h \to 0 } \frac{f(x+ h ) - f(x)}{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ U(x+h).V(x+h) - U(x).V(x) }{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ U(x+h).V(x+h) - U(x).V(x) + [U(x+h).V(x) - U(x+h).V(x) ] }{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ [ U(x+h).V(x+h) - U(x+h).V(x) ] + [ U(x+h).V(x) - U(x).V(x) ] }{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ U(x+h)[V(x+h) - V(x) ] + V(x) [ U(x+h) - U(x) ] }{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ U(x+h)[V(x+h) - V(x) ] }{h} + \frac{ V(x) [ U(x+h) - U(x) ] }{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ U(x+h)[V(x+h) - V(x) ] }{h} + \displaystyle \lim_{ h \to 0 } \frac{ V(x) [ U(x+h) - U(x) ] }{h} \\ & = \displaystyle \lim_{ h \to 0 } U(x+h) \frac{V(x+h) - V(x) }{h} + \displaystyle \lim_{ h \to 0 } V(x) \frac{ U(x+h) - U(x) }{h} \\ & = \displaystyle \lim_{ h \to 0 } V(x) \frac{ U(x+h) - U(x) }{h} + \displaystyle \lim_{ h \to 0 } U(x+h) \frac{V(x+h) - V(x) }{h} \\ & = \displaystyle \lim_{ h \to 0 } V(x) \displaystyle \lim_{ h \to 0 } \frac{ U(x+h) - U(x) }{h} + \displaystyle \lim_{ h \to 0 } U(x+h) \displaystyle \lim_{ h \to 0 } \frac{V(x+h) - V(x) }{h} \\ & = V(x) . U^\prime (x) + U(x+0) . V^\prime (x) \\ & = V(x) . U^\prime (x) + U(x) . V^\prime (x) \\ & = U^\prime (x) . V(x) + U(x) . V^\prime (x) \end{align}$
Terbukti :

$y = U.V \rightarrow y^\prime = U^\prime . V + U. V^\prime$

#Pembuktian rumus Turunan U/V
Fungsinya

$y = \frac{U}{V} \rightarrow f(x) = \frac{U(x)}{V(x)}$

$f(x+h) = \frac{U(x+h)}{V(x+h)}$
Diturunkan

$\begin{align} f^\prime (x) & = \displaystyle \lim_{ h \to 0 } \frac{f(x+ h ) - f(x)}{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ \frac{U(x+h)}{V(x+h)} - \frac{U(x)}{V(x)} }{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ \frac{V(x).U(x+h) - U(x). V(x+h) }{V(x).V(x+h)} }{h} \\ & = \displaystyle \lim_{ h \to 0 } \frac{ V(x).U(x+h) - U(x). V(x+h) }{h . V(x).V(x+h) } \\ & = \displaystyle \lim_{ h \to 0 } \frac{ V(x).U(x+h) + [ - V(x).U(x) + U(x).V(x) ] - U(x). V(x+h) }{h . V(x).V(x+h) } \\ & = \displaystyle \lim_{ h \to 0 } \frac{ [V(x).U(x+h) - V(x).U(x) ] + [ U(x).V(x) - U(x). V(x+h) ] }{h . V(x).V(x+h) } \\ & = \displaystyle \lim_{ h \to 0 } \frac{ V(x)[U(x+h) - U(x) ] + U(x)[ V(x) - V(x+h) ] }{h . V(x).V(x+h) } \\ & = \displaystyle \lim_{ h \to 0 } \frac{ \frac{V(x)[U(x+h) - U(x) ]}{h} - \frac{U(x)[ V(x+h) - V(x) ]}{h} }{V(x).V(x+h) } \\ & = \frac{ \displaystyle \lim_{ h \to 0 } \frac{V(x)[U(x+h) - U(x) ]}{h} - \displaystyle \lim_{ h \to 0 } \frac{U(x)[ V(x+h) - V(x) ]}{h} }{ \displaystyle \lim_{ h \to 0 } V(x).V(x+h) } \\ & = \frac{ \displaystyle \lim_{ h \to 0 } V(x) \displaystyle \lim_{ h \to 0 } \frac{[U(x+h) - U(x) ]}{h} - \displaystyle \lim_{ h \to 0 } U(x) \displaystyle \lim_{ h \to 0 } \frac{[ V(x+h) - V(x) ]}{h} }{ \displaystyle \lim_{ h \to 0 } V(x).V(x+h) } \\ & = \frac{ V(x) U^\prime (x) - U(x) V^\prime (x) }{ V(x).V(x+0) } \\ & = \frac{ U^\prime (x) . V(x) - U(x) . V^\prime (x) }{ V(x).V(x) } \\ & = \frac{ U^\prime (x) . V(x) - U(x) . V^\prime (x) }{ [V(x)]^2 } \end{align}$
Terbukti :