Differentiating $frac{1}{Q(x)}$ and $frac{1}{sqrt{2 pi}}e^{-frac{x^2}{2 }}$
I am not very sure about differentiating $frac{1}{Q(x)}$ and $frac{1}{sqrt{2 pi}}e^{-{x^2}/{2}}$
$$Q(x)=frac{1}{sqrt{2 pi}}int ^{infty}_{x} e^{-{t^2}/{2}} mathrm dt$$
Are my calculations below correct?
$$left(frac{1}{Q(x)}right)'= frac{1}{Q(x)^2}$$ $$left(frac{1}{sqrt{2 pi}}e^{-{x^2}/2}right)'=frac{x}{sqrt{2 pi}}e^{-{x^2}/{2}}$$
derivatives
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I am not very sure about differentiating $frac{1}{Q(x)}$ and $frac{1}{sqrt{2 pi}}e^{-{x^2}/{2}}$
$$Q(x)=frac{1}{sqrt{2 pi}}int ^{infty}_{x} e^{-{t^2}/{2}} mathrm dt$$
Are my calculations below correct?
$$left(frac{1}{Q(x)}right)'= frac{1}{Q(x)^2}$$ $$left(frac{1}{sqrt{2 pi}}e^{-{x^2}/2}right)'=frac{x}{sqrt{2 pi}}e^{-{x^2}/{2}}$$
derivatives
add a comment |
I am not very sure about differentiating $frac{1}{Q(x)}$ and $frac{1}{sqrt{2 pi}}e^{-{x^2}/{2}}$
$$Q(x)=frac{1}{sqrt{2 pi}}int ^{infty}_{x} e^{-{t^2}/{2}} mathrm dt$$
Are my calculations below correct?
$$left(frac{1}{Q(x)}right)'= frac{1}{Q(x)^2}$$ $$left(frac{1}{sqrt{2 pi}}e^{-{x^2}/2}right)'=frac{x}{sqrt{2 pi}}e^{-{x^2}/{2}}$$
derivatives
I am not very sure about differentiating $frac{1}{Q(x)}$ and $frac{1}{sqrt{2 pi}}e^{-{x^2}/{2}}$
$$Q(x)=frac{1}{sqrt{2 pi}}int ^{infty}_{x} e^{-{t^2}/{2}} mathrm dt$$
Are my calculations below correct?
$$left(frac{1}{Q(x)}right)'= frac{1}{Q(x)^2}$$ $$left(frac{1}{sqrt{2 pi}}e^{-{x^2}/2}right)'=frac{x}{sqrt{2 pi}}e^{-{x^2}/{2}}$$
derivatives
derivatives
edited Dec 2 at 11:04
amWhy
191k28224439
191k28224439
asked Dec 1 at 12:42
shineele
377
377
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add a comment |
1 Answer
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$$left( frac1{Q(x)}right)'=left((Q(x))^{-1}right)'=-Q(x)^{-2}Q'(x)$$
Try to use fundamental theorem of calculus.
$$left( frac1{sqrt{2pi}} expleft( -frac{x^2}{2}right)right)'=frac1{sqrt{2pi}} expleft( -frac{x^2}{2}right)frac{d}{dx}left( -frac{x^2}{2}right)$$
Try to differentiate again.
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$$left( frac1{Q(x)}right)'=left((Q(x))^{-1}right)'=-Q(x)^{-2}Q'(x)$$
Try to use fundamental theorem of calculus.
$$left( frac1{sqrt{2pi}} expleft( -frac{x^2}{2}right)right)'=frac1{sqrt{2pi}} expleft( -frac{x^2}{2}right)frac{d}{dx}left( -frac{x^2}{2}right)$$
Try to differentiate again.
add a comment |
$$left( frac1{Q(x)}right)'=left((Q(x))^{-1}right)'=-Q(x)^{-2}Q'(x)$$
Try to use fundamental theorem of calculus.
$$left( frac1{sqrt{2pi}} expleft( -frac{x^2}{2}right)right)'=frac1{sqrt{2pi}} expleft( -frac{x^2}{2}right)frac{d}{dx}left( -frac{x^2}{2}right)$$
Try to differentiate again.
add a comment |
$$left( frac1{Q(x)}right)'=left((Q(x))^{-1}right)'=-Q(x)^{-2}Q'(x)$$
Try to use fundamental theorem of calculus.
$$left( frac1{sqrt{2pi}} expleft( -frac{x^2}{2}right)right)'=frac1{sqrt{2pi}} expleft( -frac{x^2}{2}right)frac{d}{dx}left( -frac{x^2}{2}right)$$
Try to differentiate again.
$$left( frac1{Q(x)}right)'=left((Q(x))^{-1}right)'=-Q(x)^{-2}Q'(x)$$
Try to use fundamental theorem of calculus.
$$left( frac1{sqrt{2pi}} expleft( -frac{x^2}{2}right)right)'=frac1{sqrt{2pi}} expleft( -frac{x^2}{2}right)frac{d}{dx}left( -frac{x^2}{2}right)$$
Try to differentiate again.
edited Dec 2 at 10:51
shineele
377
377
answered Dec 1 at 12:52
Siong Thye Goh
99k1464117
99k1464117
add a comment |
add a comment |
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