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attempt 1: 10 attempts remaining. differentiate $y = (x^2 - 5)ln x$ answer: $\frac{dy}{dx} = $ submit answer next item
Step1: Identify the product rule
We have \( y=(x^{2}-5)\ln x \), which is a product of two functions \( u = x^{2}-5 \) and \( v=\ln x \). The product rule for differentiation is \( \frac{d}{dx}(uv)=u'v + uv' \).
Step2: Differentiate \( u \) and \( v \)
- Differentiate \( u = x^{2}-5 \) with respect to \( x \): \( u'=\frac{d}{dx}(x^{2}-5)=2x \)
- Differentiate \( v=\ln x \) with respect to \( x \): \( v'=\frac{d}{dx}(\ln x)=\frac{1}{x} \)
Step3: Apply the product rule
Using the product rule \( \frac{dy}{dx}=u'v + uv' \), substitute \( u = x^{2}-5 \), \( u' = 2x \), \( v=\ln x \), and \( v'=\frac{1}{x} \):
\[
$$\begin{align*}
\frac{dy}{dx}&=2x\cdot\ln x+(x^{2}-5)\cdot\frac{1}{x}\\
&=2x\ln x+\frac{x^{2}-5}{x}\\
&=2x\ln x + x-\frac{5}{x}
\end{align*}$$
\]
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\( 2x\ln x + x-\frac{5}{x} \)