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Differentiate, with respect to $x$, $\frac{e^{-2x}}{\sqrt x}$.

I a having difficulties with differentiation. The answer is

$$-\frac{e^{-2x}(4x+1)}{2x\sqrt{x}}$$

But I am looking at my working out and can't seem to solve the question. This is what I did:

$$\frac{d}{dx}\left(\frac{e^{-2x}}{\sqrt{x}}\right) = \frac{\sqrt{x}\times \frac{d}{dx}(e^{-2x})-e^{-2x}\times \frac{d}{dx}(\sqrt{x})}{(\sqrt{x})^2} = \frac{\sqrt{x}(-2e^{2x})-e^{-2x}\frac12(\sqrt{x})^{-1/2}}{x}$$ $$ = \frac{\sqrt{x}(-2e^{2x})-e^{-2x}}{2x\sqrt{x}}$$

image of my work

If someone could please help me solve this question or give advise please help me! Thank You!

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    $\begingroup$ The expression at the end of line $2$ has some mistakes in numerator: $(1)$ you missed minus sign of exponent of $e$. $(2)$ the last expression should be $\frac12x^{\frac{-1}2}$ not $\frac12(\sqrt x) ^{\frac{-1}2}$. $\endgroup$ – Soheil Feb 4 at 8:08
  • $\begingroup$ To make things a little easier, rewrite the given expression as $-\frac 12e^{-2x}x^{-3/2}(4x+1)$ and use the product rule. $\endgroup$ – Karl Feb 4 at 8:08
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    $\begingroup$ I have restored errors in the original mary james' attempt, based on the image linked. $\endgroup$ – CiaPan Feb 4 at 8:53
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    $\begingroup$ @Karl This will simplify the problem, of course, but it looks like the task was explicitly to use a quotient rule (see the image linked). $\endgroup$ – CiaPan Feb 4 at 8:55
  • $\begingroup$ i seee thank you so much for the help! $\endgroup$ – mary james Feb 4 at 9:01
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We have $f(x)=\dfrac{e^{-2x}}{\sqrt x}$. Use Quotient rule as you suggested in your answer too: $$f'(x)=\frac{-2e^{-2x}\sqrt x -\frac{1}{2\sqrt x}\times e^{-2x}}{x}$$ Now multiply the fraction by $\frac{2\sqrt x}{2\sqrt x}$ to get rid of $\frac{-1}{2\sqrt x}$ in numerator:

$$\frac{-2e^{-2x}\sqrt x -\frac{1}{2\sqrt x}\times e^{-2x}}{x}\times\frac{2\sqrt x}{2\sqrt x}=\frac{-4e^{-2x}\times x-e^{-2x} }{2x\sqrt x}=-\frac{e^{-2x}(4x+1)}{2x\sqrt x}$$

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  • $\begingroup$ How do you get rid of the √x by multiplying it by 2√x/2√x? $\endgroup$ – mary james Feb 4 at 9:10
  • $\begingroup$ @maryjames Consider $\sqrt x\times\sqrt x = x$ and $\sqrt x\times \frac{1}{\sqrt x}=1$. When you multiply the fraction by $\frac{2\sqrt x}{2\sqrt x}$ it is the same as multiplying numerator by $2\sqrt x$ and denominator by $2\sqrt x$. $\endgroup$ – Soheil Feb 4 at 9:12
  • $\begingroup$ I understand how you got rid of -1/2√x but the one with -2e^-2x√x how did the √x disappear $\endgroup$ – mary james Feb 4 at 9:15
  • $\begingroup$ $(-2e^{-2x}\sqrt x)\times 2\sqrt x=-4e^{-2x}\times \sqrt x\times \sqrt x$. $\endgroup$ – Soheil Feb 4 at 9:16
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    $\begingroup$ No problem! Glad to have helped you. $\endgroup$ – Soheil Feb 4 at 9:19
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Note that $\frac{1}{\sqrt x} = x^{-1/2}$. Now use the product rule:

$$\frac{d}{dx} \left(e^{-2x} \cdot x^{-1/2} \right)= e^{-2x} \cdot -\frac{1}{2}x^{-3/2} + (e^{-2x} \cdot -2) \cdot x^{-1/2}$$

which is already correct, but can be written in a nicer form: $$=e^{-2x} \left(-\frac{1}{2}x^{-3/2} - 2x^{-1/2} \right)=e^{-2x} x^{-3/2} \left(-\frac{1}{2} - 2x \right)= -\frac{1/2 + 2x}{e^{2x} x^{3/2}} = -\frac{1+4x}{2e^{2x} x^{3/2}}$$

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You need to use $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{d}{dx}u-u\frac{d}{dx}v}{v^2}.$$

When this formula is compared to your question, you have $u = e^{-2x}$ and $v=\sqrt{x}.$ Just substitute and simplify.

That is $$\frac{d}{dx}\left(\frac{e^{-2x}}{\sqrt{x}}\right) = \frac{\sqrt{x}\frac{de^{-2x}}{dx}-e^{-2x}\frac{d\sqrt{x}}{dx}}{(\sqrt{x})^2}.$$

$$\frac{d}{dx}\left(\frac{e^{-2x}}{\sqrt{x}}\right) = \frac{\sqrt{x}e^{-2x}(-2)-e^{-2x}\frac{1}{2\sqrt{x}}}{(\sqrt{x})^2}.$$

After simplification, one should get

$$\frac{d}{dx}\left(\frac{e^{-2x}}{\sqrt{x}}\right) = \frac{-4xe^{-2x}-e^{-2x}}{2x\sqrt{x}} = -\frac{e^{-2x}(4x+1)}{2x\sqrt{x}} $$

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