Y = sec θ tan θ differentiate

1 answer:

6 0

Y = sec t tan t = 1/ cos t * sin t / cos t = sin t / cos² t
$y`= \frac{cos t cos ^{2} t-2 cost*(-sin t)*sint}{cos^{4} t}= \\ =\frac{cos ^{3} t+2costsin^{2}t }{cos ^{4}t }= \\ = \frac{cos ^{2} t+2sin ^{2} t}{cos ^{3} t}= \\ = \frac{1-sin ^{2}t+2sin ^{2} t }{cos ^{3}t }= \\ = \frac{1+sin ^{2} t}{cos ^{3}t }$

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Given : In a hypothetical study looking at the scores of 100 people in an entrance exam, it was determined that the average score was 95 with a standard deviation of 10.

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