Jawaban:

Turunan fungsi y = tan x adalah sec² x

Pembahasan

TRIGONOMETRI

Pada sebuah segitiga siku siku ABC , perhatikan lampiran, pada sudut α,

sisi di samping sudut AB = x

sisi di depan sudut BC = y

sisi miring AC = r

r² = x² + y²

Berlaku

\begin{gathered}sin \: \alpha \:=\: \frac{y}{r}\\cos \:\alpha \:=\: \frac{x}{r}\\tan \: \alpha \:=\: \frac{y}{x}\\sec \: \alpha \:=\: \frac{r}{x}\\cosec \: \alpha \:=\: \frac{r}{y}\\cotan \: \alpha \:=\: \frac{x}{y}\end{gathered}

sinα=

r

y

cosα=

r

x

tanα=

x

y

secα=

x

r

cosecα=

y

r

cotanα=

y

x

Dari rumus diatas, dapat disimpulkan

\begin{gathered}cosec\: \alpha \:=\: \frac{1}{sin \: \alpha }\\sec \: \alpha \:=\: \frac{1}{cos \: \alpha }\\tan \: \alpha \:=\: \frac{sin \: \alpha }{cos \: \alpha }\\cotan \: \alpha \:=\:\frac{cos \: \alpha }{sin \: \alpha }\end{gathered}

cosecα=

sinα

1

secα=

cosα

1

tanα=

cosα

sinα

cotanα=

sinα

cosα

r² = x² + y²

Semua bagi dengan r², menjadi

\begin{gathered}\frac{r^2}{r^2} \:=\: \frac{x^2}{r^2} \:+\: \frac{y^2}{r^2}\\1\:=\: (\frac{x}{r})^2 \:+\: (\frac{y}{r})^2\\1 \:=\: (sin\: x)^2 \:+\: (cos \: x)^2\end{gathered}

r

2

r

2

=

r

2

x

2

+

r

2

y

2

1=(

r

x

)

2

+(

r

y

)

2

1=(sinx)

2

+(cosx)

2

sin² x + cos² x = 1

Turunan Aljabar

Rumus - rumus turunan aljabar

1. f (x) = k ⇒ f' (x) = 1

2. f (x) = axⁿ ⇒ f' (x) = an xⁿ⁻¹

3. f (x) = g (x) ± h(x) ⇒ f' (x) = g' (x) ± h' (x)

4. f (x) = g(x) . h(x) ⇒ f' (x) = g' (x) h(x) + g (x) h' (x)

5. f(x) \:=\: \frac{g(x)}{h(x)}f(x)=

h(x)

g(x)

⇒ f'(x) \:=\: \frac{g'(x) \: h(x) \:-\: g(x) \: h'(x)}{h^2(x)}f

′

(x)=

h

2

(x)

g

′

(x)h(x)−g(x)h

′

(x)

Turunan Sin x

Definisi turunan bisa diperoleh dari limit

\begin{gathered}d(sin\:x) \:=\:\lim_{h \to 0} \frac{sin\: (x\:+\: h) \:-\: sin \: x}{h}\\=\: \lim_{h \to 0} \frac{sin\: x \: cos \: h \:+\: cos \: x \: sin \: h \: -\: sin \: x}{h}\end{gathered}

d(sinx)=

h→0

lim

h

sin(x+h)−sinx

=

h→0

lim

h

sinxcosh+cosxsinh−sinx

\begin{gathered}=\: \lim_{h \to 0} (- \: sin \: x\: (\frac{1 \:-\: cos \: h}{h}) \:+\: cos \: x\: (\frac{sin \: h}{h} ))\\=\: (-\: sin\:x) \: [\lim_{h\to 0} \frac{1\:-\: cos \:h}{h}] \:+\: (cos \:x) \: [\lim_{h \to 0} \frac{sin \: h}{h}]\end{gathered}

=

h→0

lim

(−sinx(

h

1−cosh

)+cosx(

h

sinh

))

=(−sinx)[

h→0

lim

h

1−cosh

]+(cosx)[

h→0

lim

h

sinh

]

= (- sin x) [0] + cos x [1]

= cos x

Turunan cos x

\begin{gathered}d(cos\:x) \:=\: \lim_{h \to 0} \frac{cos\: (x\:+\: h) \:-\: cos \: x}{h}\\=\: \lim_{h \to 0} \frac{cos \: x \: cos \: h \:-\: sin\: x \: sin \: h \: -\: cos \: x}{h}\end{gathered}

d(cosx)=

h→0

lim

h

cos(x+h)−cosx

=

h→0

lim

h

cosxcosh−sinxsinh−cosx

\begin{gathered}=\: \lim_{h \to 0} (- \: cos \: x\: (\frac{1 \:-\: cos \: h}{h}) \:-\: sin\: x\: (\frac{sin \: h}{h} ))\\=\: (-\: cos\:x) \: [\lim_{h\to 0} \frac{1\:-\: cos \:h}{h}] \:-\: (sin\:x)\:[\lim_{h \to 0}\frac{sin\:h}{h}]\end{gathered}

=

h→0

lim

(−cosx(

h

1−cosh

)−sinx(

h

sinh

))

=(−cosx)[

h→0

lim

h

1−cosh

]−(sinx)[

h→0

lim

h

sinh

]

= (- cos x) [0] - sin x [1]

= - sin x

Dit:

Turunan tan x ?

Penjelasan:

d tan x

=\: d(\frac{sin \:x}{cos \: x})=d(

cosx

sinx

)

Uraikan sesuai rumus turunan aljabar no.5

=\: \frac{d(sin \:x) \: cos \: x \:-\: sin\:x \: d(cos \: x)}{cos^2 \: x}=

cos

2

x

d(sinx)cosx−sinxd(cosx)

=\: \frac{cos \: x \: cos \: x \:-\: sin \: x \: (- \: sin \: x)}{cos ^2 \: x}=

cos

2

x

cosxcosx−sinx(−sinx)

=\: \frac{cos^2 \: x \:+\: sin^2 \: x}{cos^2\: x}=

cos

2

x

cos

2

x+sin

2

x

\begin{gathered}=\: \frac{1}{cos^2 \: x}\\=\: (\frac{1}{cos \:x})^2\end{gathered}

=

cos

2

x

1

=(

cosx

1

)

2

Karena sec x = \frac{1}{cos\:x}

cosx

1

= (sec x)²

= sec² x

Penjelasan dengan langkah-langkah:

f(x)=tan2x

f'(x)=2sec²2x

f"(x)=4sec2x(2sec2x.tan2x)

=8sec²2x.tan2x