## Related questions with answers

Find the derivatives of the functions in the ff exercise

$k(x)=x^2 \sec (\frac{1}{x})$

Solutions

VerifiedWe are required to obtain the first derivative of the given function $k=k(x).$ This function contains the product of an algebraic function and a composite function, which will be the most challenging part to differentiate.

In this exercise we use the following formulas

$(1)\color{#c34632}{\,\,\, \dv{x} (x^n)=nx^{n-1}}$

$(2)\color{#c34632}{\,\,\, \dv{x} ( uv)=v \dv{u}{x}+u\dv{v}{x}}$

$(3)\color{#c34632}{\,\,\, (f \circ g)'(x)=f'\left(g(x)\right)g'(x)}$

In this case, it is

$u=x^2 \text{ and } v=\sec{\left(\frac 1x \right)}$

Now, we obtain

$\begin{align*} k'&\mathop=\limits^{(1),(2)} \sec{\left(\frac 1x \right)}\cdot 2x+x^2 \left[\sec{\left(\frac 1x \right)} \right]'\\ &\mathop=\limits^{(3)} 2x \sec{\left(\frac 1x \right)} +x^2 \left[\sec{\left(\frac 1x \right)} \tan {\left(\frac 1x \right)}\left(\frac 1x \right)' \right]\\ &\mathop=\limits^{(1)} 2x \sec{\left(\frac 1x \right)} +x^2\sec{\left(\frac 1x \right)} \tan {\left(\frac 1x \right)} \cdot \left( -\frac {1}{x^2} \right)\\ &=2x \sec{\left(\frac 1x \right)}-\sec{\left(\frac 1x \right)} \tan {\left(\frac 1x \right)}\\ &=\sec{\left(\frac 1x \right)} \left[ 2x-\tan {\left(\frac 1x \right)} \right] \end{align*}$

$k(x)=x^2\sec(\dfrac{1}{x})$

$k'(x)=(x^2)'\cdot \sec (\dfrac{1}{x})+x^2\cdot (\sec(\dfrac{1}{x}))'=\\\\ =2x\sec (\dfrac{1}{x})+x^2\cdot \sec(\dfrac{1}{x})\tan (\dfrac{1}{x})\cdot (\dfrac{1}{x})'=\\\\ =2x\sec (\dfrac{1}{x})-x^2 \sec(\dfrac{1}{x})\tan (\dfrac{1}{x})\cdot x^{-2}=\\\\ =2x\sec (\dfrac{1}{x})-\sec(\dfrac{1}{x})\tan (\dfrac{1}{x})$

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