53 lines
1.4 KiB
TeX
53 lines
1.4 KiB
TeX
\begin{enumerate}[label=\arabic*.]
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\item \[ f(x) = ln\left(\frac{1}{x^2}\right) \]
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La $u = \frac{1}{x^2}$
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Jeg bruker kjerneregelen:
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\begin{align*}
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\frac{dy}{dx} &= \frac{dy}{du} \cdot \frac{du}{dx} \\
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&= \frac{d}{du} ln(u) \cdot \frac{d}{dx} \frac{1}{x^2} \\
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&= \frac{1}{u} -2 \frac{1}{x^3} \\
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&= \frac{1}{\frac{1}{x^2}} -2 \frac{1}{x^3} \\
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&= x^2 -2 \frac{1}{x^3}
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\end{align*}
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\item \[g(x) = \frac{1 + \sin x}{1 + e^x + x^2}\]
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Jeg bruker kvotientregelen:
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$u = 1 + \sin x$
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$v = 1 + e^x + x^2$
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\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{ \frac{d}{dx}(u) \cdot v - u \cdot \frac{d}{dx}(v)}{v^2} \]
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\begin{align*}
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\frac{d}{dx} u &= \frac{d}{dx} 1 + \sin x \\
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&= \cos x
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\end{align*}
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\begin{align*}
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\frac{d}{dx} v &= \frac{d}{dx} 1 + e^x + x^2 \\
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&= e^x + 2x
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\end{align*}
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\begin{align*}
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\frac{dg}{dx} &= \frac{(\cos x)(1+e^x+x^2) - (1+\sin x)(e^x + 2x)}{\left(1 + e^x + x^2\right)^2}
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\end{align*}
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\item \[h(x) = \sqrt{1 + \sqrt{x}}\]
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$u = 1 + \sqrt{x}$
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\begin{align*}
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\frac{dy}{dx} &= \frac{dy}{dv} \cdot \frac{du}{dx} \\
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&= \frac{d}{du} \sqrt{u} \cdot \frac{d}{dx} \left( 1 + \sqrt{x} \right) \\
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&= \frac{1}{2\sqrt{u}} \cdot \frac{1}{2\sqrt{x}} \\
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&= \frac{1}{2\sqrt{1+\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \\
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&= \frac{1}{4\sqrt{1+\sqrt{x}} \sqrt{x}}
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\end{align*}
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\end{enumerate} |