1.
$y=\tan(\ln{x})$ maka turunannya adalah ….
Jawab:
Ini menggunakan aturan rantai seperti: $f(x)=\tan(g(x))$ maka
$f’(x)=\sec^2(g(x))\times g’(x)$ sehingga
$y=\tan(\ln{x})$ maka $\frac{dy}{dx}=\sec^2(\ln{x})\times \frac{1}{x}$
$\frac{dy}{dx}=\frac{1}{x}\sec^2(\ln{x})$
2.
$y=\ln{\arctan(\tan(\frac{x}{2}))}$
Jawab:
Misalkan
$a=\tan(\frac{x}{2})\text{ maka }\frac{da}{dx}=\frac{d(\tan(\frac{x}{2})}{dx}=\frac{1}{2}\sec^2(\frac{x}{2})$
$b=\arctan(a)\text{ maka
} \frac{db}{da}=\frac{d(\arctan(a))}{da}=\frac{1}{a^2+1}$
Maka,
$\frac{dy}{dx}=\frac{d}{dx}\left[\ln{b}\right]$
$\frac{dy}{dx}=\frac{da}{dx}\times\frac{db}{da}\times\frac{d}{db}\left[\ln{b}\right]$
$\frac{dy}{dx}=\frac{1}{2}\sec^2(\frac{x}{2})\times\frac{1}{a^2+1}\times\frac{1}{b}$
$\frac{dy}{dx}=\frac{\sec^2(\frac{x}{2})}{2(a^2+1)(b)}$
$\frac{dy}{dx}=\frac{\sec^2(\frac{x}{2})}{2\left(\tan^2(\frac{x}{2}) +1\right)(\arctan(a))}$
$\frac{dy}{dx}=\frac{\sec^2(\frac{x}{2})}{2\left(\tan^2(\frac{x}{2}) +1\right)\left(\arctan(tan(\frac{x}{2}))\right)}$
$\frac{dy}{dx}=\frac{\sec^2(\frac{x}{2})}{2\left(\tan^2(\frac{x}{2}) +1\right)\left(\arctan(\tan(\frac{x}{2}))\right)}$
3.
$y=\arctan(\ln{\sin(x)})$
Jawab:
$\frac{dx}{dy}=\frac{d}{dx}\left[\arctan(\ln{\sin(x)})\right]$
Misalkan:
$a=\sin(x) \text{ maka } \frac{da}{dx}=\frac{d(\sin(x))}{dx}=\cos(x)$
$b=\ln{a} \text{ maka }
\frac{db}{da}=\frac{1}{a}=\frac{1}{\sin(x)}$
Maka;
$\frac{dx}{dy}=\frac{d}{dx}\left[\arctan(\ln{sin(x)})\right]$
$\frac{dx}{dy}=\frac{da}{dx}\times \frac{db}{da}\frac{d(arctan(b))}{db}$
$\frac{dx}{dy}= (\cos(x))\left(\frac{1}{\sin(x)}\right)\left(\frac{1}{b^2+1}\right)$
$\frac{dx}{dy}=\frac{\cos(x)}{(\sin(x))((\ln{a})^2+1)}$
$\frac{dx}{dy}=\frac{\cos(x)}{(\sin(x))((\ln{sin(x)})^2+1)}$
4.
$y=\ln{e^{7x^2-x+1}}$
Jawab:
$y=\ln{e^{7x^2-x+1}}$
$y= (7x^2-x+1)\ln{e}$
$y=7x^2-x+1$
$\frac{dy}{dx}=\frac{d(7x^2-x+1)}{dx}$
$\frac{dy}{dx}=\frac{d(7x^2-x+1)}{dx}$
$\frac{dy}{dx}=14x-1$
5.
$y=e.a^{x/e}$
Jawab:
Cara 1:
$y=e.a^{x/e}$
Misal: $u=e$ maka $\frac{du}{dx}=0$
$v=a^{x/e}$ maka $\frac{dv}{dx}=\frac{1}{e}a^{x/e}\ln{a}$
$y=u.v$ maka $\frac{dy}{dx}=v.\frac{du}{dx}+u.\frac{dv}{dx}$
$\frac{dy}{dx}=a^{x/e}.0+e. \frac{1}{e}a^{x/e}\ln{a}$
$\frac{dy}{dx}= a^{x/e}\ln{a}$
Cara 2:
$y=e.a^{x/e}$
$\ln{y}=\ln{e.a^{x/e}}$
$\ln{y}=\ln{e} + \ln{a^{x/e}}$
$\ln{y}= 1+ x/e \ln{a}$
$\frac{d(\ln{y})}{dx}=\frac{d(1)}{dx}+\frac{d(x/e\ln {a})}{dx}$
$\frac{1}{y} \frac{dy}{dx}=0+\frac{\ln{a}}{e}$
$\frac{dy}{dx}=y.\frac{\ln{a}}{e}$
$\frac{dy}{dx}=e.a^{x/e}/\frac{\ln{a}}{e}$
$\frac{dy}{dx}=a^{x/e}\ln{a}$
6.
$y=x^n+n^x$ (n konstanta)
Jawab:
$y=x^n+n^x$ maka $\frac{dy}{dx}=\frac{d(x^n)}{dx}+\frac{d(n^x)}{dx}$
$\frac{dy}{dx}=nx^{n-1}+n^x\ln{n}$
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