Tentukan Turunan Pertama Dari :.... Question from @MasKibo

May 2020 1 177 Report

Tentukan Turunan Pertama Dari :

$a. \: y = \sqrt{ {x}^{2} \: cos \: x }$
$b. \: cos(xy) = {y}^{2} + 2x$

Untuk soal yang a.

$y=\sqrt{x^2\cos x}\\\\y=x\sqrt{\cos x}\\\\y=x(\cos x)^{\frac{1}{2}}\\\\--------------\,\frac{d}{dx}(...)\\\\\frac{d}{dx}(y)=\frac{d}{dx}[x(\cos x)^{\frac{1}{2}}]\\\\\frac{d}{dx}(y)=(\cos x)^{\frac{1}{2}}\frac{d}{dx}(x)+x\frac{d}{dx}(\cos x)^{\frac{1}{2}}\\\\\frac{d}{dx}(y)=(\cos x)^{\frac{1}{2}}\frac{d}{dx}(x)+x[\frac{1}{2}(\cos x)^{-\frac{1}{2}}](-\sin x)\\\\\frac{d}{dx}(y)=(\cos x)^{\frac{1}{2}}+x[\frac{1}{2\sqrt{\cos x}}](-\sin x)\\\\\frac{d}{dx}(y)=\sqrt{\cos x}-\frac{x\sin x}{2\sqrt{\cos x}}$

$\frac{d}{dx}(y)=\frac{2\cos x}{2\sqrt{\cos x}}-\frac{x\sin x}{2\sqrt{\cos x}}\\\\\frac{d}{dx}(y)=\frac{2\cos x-x\sin x}{2\sqrt{\cos x}}$

Untuk soal yang b.

$\cos(xy)=y^2+2x\\\\--------------\,\frac{d}{dx}(...)\\\\\frac{d}{dx}[\cos(xy)]=\frac{d}{dx}(y^2+2x)\\\\\{-\sin(xy)\}\frac{d}{dx}(xy)=\frac{d}{dx}(y^2)+\frac{d}{dx}(2x)\\\\\{-\sin(xy)\}(y+x\frac{dy}{dx})=2y\frac{dy}{dx}+2\\\\-y\sin(xy)-x\sin(xy)\frac{dy}{dx}=2y\frac{dy}{dx}+2\\\\-2y\frac{dy}{dx}-x\sin(xy)\frac{dy}{dx}=2+y\sin(xy)\\\\\--[2y+x\sin(xy)]\frac{dy}{dx}=2+y\sin(xy)\\\\\frac{dy}{dx}=-\frac{2+y\sin(xy)}{2y+x\sin(xy)}$

Sekian dulu untuk jawaban saya ini. Mohon untuk dikoreksi kembali jawaban saya ini. Terimakasih :)

MasKibo kak tolong satu lagi dong

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