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Recently, I have been watching Professor David Jerison's Lectures on Calculus 1 on MIT's Open Course Ware website. On the topic of linear and quadratic approximations, The following examples have been brought:

Linear approx of $e^x$ near $x=0$ is $x+1$; and Linear approx of $(x+1)^r$ near $x=0$ is $rx+1$.

Then, the professor proceeds to say that the linear approximation of $e^{-3x}(x+1)^{-{1\over2}}$ would, applying the previously mentioned approximations, be simply $(1-3x)(1-{1\over2}x)$.

My question is, why can we individually approximate a constituent part of the function, then multiply the result? It seems to me that since ${{d\over dx}uv} \neq ({d\over dx} u)({d\over dx}v)$ where $u$ and $v$ are functions of $x$, then we should not be able to simply multiply each individual approximation? This is especially since the formula for linear approximations require taking the derivative (the approximation formula is $f(x)+f'(x)(x-x_0)$.

Thanks in advance.

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    $\begingroup$ I wouldn't say that $(1-3x)(1-x/2)$ was a linear approximation. $\endgroup$ – Lord Shark the Unknown Sep 9 '17 at 4:37
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As @LordSharkTheUnknown pointed out in comments, $(1-3x)(1-x/2)$ is not exactly linear since it has a quadratic term in in. But the linear approximation to $e^{-3x}(x+1)^{-1/2}$ can be obtained from it if we multiply it out and ignore higher-degree terms, i.e. the quadratic term in this case: $$e^{-3x}(x+1)^{-1/2}\approx(1-3x)\left(1-\frac{x}{2}\right)=1-3x-\frac{x}{2}+\frac{3x^2}{2}=1-\frac{7x}{2}+O(x^2).$$ The logic behind such multiplication is elementary: if $A=B$ and $C=D$, then $AC=BD$, isn't it?

Moreover, thanks to the distributive properties, this multiplication in fact produces the correct Product Rule and in a sense explains it!

Let $f(x)\approx f(0)+f'(0)x$ and $g(x)\approx g(0)+g'(0)x$ be the linear approximations, a.k.a. the differentials, of two differentiable functions around $x_0=0$. Then: $$f(x)g(x)\approx(f(0)+f'(0)x)(g(0)+g'(0)x)=f(0)g(0)+f'(0)g(0)x+f(0)g'(0)x+f'(0)g'(0)x^2=f(0)g(0)+\color{magenta}{\underbrace{\left[f'(0)g(0)+f(0)g'(0)\right]}_{(fg)'(0)}}x+O(x^2).$$

Say in your example it was the middle terms: $$e^{-3x}(x+1)^{-1/2}\approx(1-3x)\left(1-\frac{x}{2}\right)=1+\color{magenta}{\underbrace{1\cdot(-3x)+1\cdot\left(-\frac{x}{2}\right)}_{v\frac{du}{dx}+u\frac{dv}{dx}}}+\frac{3x^2}{2}=1-\frac{7x}{2}+O(x^2).$$

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$(1-3x)(1-\frac 12x)$ is not a linear approximation because it has a quadratic term, but if you just keep the first two terms you do have the linear approximation to the product as $1-\frac 72x$. $1-3x$ and $1-\frac 12x$ are the first two terms in the Taylor series of the two functions. When you multiply the Taylor series, you get the Taylor series of the product. The higher terms of the Taylor series of the individual functions cannot contribute to the constant and linear terms of the product. If you wanted a quadratic approximation to the product you would have to find the quadratic approximation to each function, multiply them, and throw away the cubic and quartic terms.

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