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jolli1 [7]
3 years ago
12

HELP ASAP I DONT HAVE TIME AND IT ALSO DETECTS IF ITS RIGHT OR WRONG

Mathematics
1 answer:
mamaluj [8]3 years ago
7 0

Answer:

75%

Step-by-step explanation:

(5.6-3.2)/3.2=0.75

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the following are scores of students:9,7,8,9,7,5,8,10,7,8 and 10. the value of the population variance is 4. true or false?​
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3 years ago
Find the surface area of the solid figure
Helen [10]

Answer:

150

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5 0
3 years ago
(15 Points)
expeople1 [14]

ANSWER 1



Note that,


f(u)=tan^{-1}(u)


is the same as


f(u)=arctan(u)



We apply the product rule.


f(x)=x^2tan^{-1}(x)


So we keep the second function and differentiate the first,plus we keep the first function and differentiate the second.


f'(x)=(x^2)'tan^{-1}(x)+x^2(tan^{-1}(x))'



Recall that,

If

f(u)=tan^{-1}(u)



Then,

f'(u)=\frac{1}{1+u^2}} \times u'


This implies that,

f'(x)=2xtan^{-1}(x)+\frac{x^2}{x^2+1}



ANSWER 2


We apply the product rule and the chain rules of differentiation here.



f(x)=xsin^{-1}(1-x^2)




f'(x)=x'sin^{-1}(1-x^2)+x(sin^{-1}(1-x^2))'



Recall that,

If

f(u)=sin^{-1}(u)



Then,

f'(u)=\frac{1}{\sqrt{1-u^2}} \times u'



This implies that,


f'(x)=sin^{-1}(1-x^2)+x \times \frac{1}{\sqrt{1-(1-x^2)^2}}\times (-2x)


f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-(1-2x^2+x^4)}}


f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-1+2x^2-x^4}}



f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{2x^2-x^4}}





5 0
4 years ago
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