Derivative of sum of two functions
I have to find $\frac{dy}{dx}\left[(x\sqrt{x})+\frac{1}{x^2\sqrt{x}}\right]$ but would like to find where I made a mistake in my solution.
Here is my work:\begin{align} (f+g)'=& \ f'+g', f = x \sqrt{x}, g= \frac{1}{x^2 \sqrt x}\\ f=& \ x \cdot x^{1/2} = x^{3/2} \\ f'=& \ \frac{3}{2x\sqrt{x}} \\ \\ g =& \ x^{-2}\cdot x^{-1/2} \\ g =& \ x^{-5/2} \\ g' =& \ -\frac{5}{2x^3 \sqrt{x}} \\ \\ f'+g' =& \ \frac{3}{2x\sqrt{x}} - \frac{5}{2x^3 \sqrt{x}} \\ =& \ \frac{3x^2-5}{2x^3 \sqrt x} \\ \frac{dy}{dx}=& \ \frac{\sqrt x (3x^2-5)}{2x^4} \end{align}
$\endgroup$ 22 Answers
$\begingroup$$$\frac{d}{dx}\left[(x\sqrt{x})+\frac{1}{x^2\sqrt{x}}\right]$$$$=\frac{d}{dx}\left(x\sqrt{x}\right)+\frac{d}{dx}\ \left(\frac{1}{x^{2}\sqrt{x}}\right)$$$$=\frac{d}{dx}\left(x^{\frac{3}{2}}\right)+\frac{d}{dx}\left(x^{\large-\frac{5}{2}}\right)$$$$=\frac{3}{2}\large x^{\frac{1}{2}}-\frac{5}{2}\large x^{-\frac{7}{2}}$$$$=\frac{3}{2}\sqrt{x}-\frac{5}{2\sqrt{x^{7}}}$$
In most high school calculus books the power rule for derivatives is proven used binomial theorem, but binomial theorem is not helpful for all arbitrary real powers.
For more information about the derivation of the function $x^{n}$ for all real $n,x$ refer to the link
$\endgroup$ $\begingroup$In general if $f(x)=x^{n}$ for some constant $n$ then $f'(x)=nx^{n-1}$ which is the power rule.
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