Turunan Pertama dari fungsi f(x) = sin²(6x) + adalah…​

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Turunan Pertama dari fungsi f(x) = sin²(6x) + adalah…​

Jawab:

f'(x)=12sin(6x)cos(6x)+dfrac{1}{2sqrt{x}}

atau

f'(x)=6sin(12x)+dfrac{1}{2sqrt{x}}

PEMBAHASAN

Turunan

f(x)=sin^2(6x)+sqrt{x}

Cara Pertama

begin{aligned}f'(x)&=left(sin^2(6x)+sqrt{x}right)'\&quadleft[ begin{aligned} &cos(2alpha)=1-2sin^2(alpha)\&{Rightarrow }sin^2(alpha)=frac{1}{2}-frac{1}{2}cos(2alpha)\&{Rightarrow }sin^2(6x)=frac{1}{2}-frac{1}{2}cos(12x)end{aligned}right.end{aligned}

begin{aligned}{Leftrightarrow }f'(x)&=left(frac{1}{2}-frac{1}{2}cos(12x)+sqrt{x}right)'\{Leftrightarrow }f'(x)&=left(frac{1}{2}right)'-frac{1}{2}left(cos(12x)right)'+left(sqrt{x}right)'\{Leftrightarrow }f'(x)&=0-frac{1}{2}left(-sin(12x)right)left(12xright)'+frac{1}{2sqrt{x}}\{Leftrightarrow }f'(x)&=frac{1}{2}sin(12x)(12)+frac{1}{2sqrt{x}}\{Leftrightarrow }f'(x)&=6sin(12x)+frac{1}{2sqrt{x}}\end{aligned}

Cara Kedua

begin{aligned}f(x)&=sin^2(6x)+sqrt{x}\\f'(x)&=left(sin^2(6x)+sqrt{x}right)'\{Leftrightarrow }f'(x)&=left(sin^2(6x)right)'+left(sqrt{x}right)'\&quadleft[ begin{aligned} &g=u^2,, u=sin(6x)\&{Rightarrow }g'=frac{dg}{du}cdotfrac{du}{dx}\&{Rightarrow }g'=left(u^2right)'left(sin(6x)right)'\&{Rightarrow }g'=2ucdotcos(6x)(6x)'\&{Rightarrow }g'=12sin(6x)cos(6x)\end{aligned}right.end{aligned}

begin{aligned}{Leftrightarrow }f'(x)&=12sin(6x)cos(6x)+frac{1}{2sqrt{x}}\{Leftrightarrow }f'(x)&=6cdot2sin(6x)cos(6x)+frac{1}{2sqrt{x}}\{Leftrightarrow }f'(x)&=6sin(12x)+frac{1}{2sqrt{x}}\end{aligned}

KESIMPULAN

∴  Turunan pertama f(x) adalah:

f'(x)=12sin(6x)cos(6x)+dfrac{1}{2sqrt{x}}

atau

f'(x)=6sin(12x)+dfrac{1}{2sqrt{x}}