Differentiate $\sqrt{x} e^x \sec x$

calculusderivatives

Differentiate the following equation : $$\sqrt{x} e^x \sec x$$

Here I can see 3 values.

  1. $\sqrt{x}$
  2. $e^x$
  3. $\sec x$

This time, how can I use product rule?

My book said

$$\sqrt{x} \frac{d}{dx}(e^x \sec (x)) + e^x \sec {x} \frac{d}{dx} (\sqrt{x})$$

But, I found three values. So, according to product rule if I differentiate than, I would have 3 variable, wouldn't I? What am I missing?

$$\frac{d}{dx}ab = a\frac{d}{dx}b + b\frac{d}{dx}a$$

It's for two values (a,b). So, It would have 3 variables for 3 values(a,b,c).

Best Answer

The product rule, which I'll write as $(fg)'=f'g+fg'$, works recursively. That is, if you have a product of three functions, for example, then

$$(fgh)'=(f(gh))'=f'(gh)+f(gh)'=f'(gh)+f(g'h+gh')=f'gh+fg'h+fgh'$$

In your specific case, you have $f(x)=\sqrt x$, $g(x)=e^x$, and $h(x)=\sec x$, so you just need to differentiate each of those functions separately and then fit the results into the final formula for the derivative of the product.

Remark: The use of parentheses here is potentially confusing but hopefully clear from context: the parentheses in $f(gh)$ and elsewhere in the displayed equation are purely for multiplicative grouping, whereas the parentheses in $f(x)=\sqrt x$ (and likewise for $g(x)$ and $h(x)$) are part of defining $f$, $g$ and $h$ as functions of $x$.

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