Can someone help me on question 9 & 10. One Is just fine. Thank you
1 answer:
Answer:
9. y = x - 1/2 + π/4
10. y = (-2/3)(√6)x + (2/3)√3 + π/3
Step-by-step explanation:
9. From a table of derivatives,
... arctan'(x) = 1/(1+x²)
Then, using the chain rule, ...
... f'(x) = arctan'(2x) = 2/(1 +(2x)²) = 2/(1 +4x²)
For x=1/2, the slope of the tangent is ...
... f'(1/2) = 2/(1 +4·(1/2)²) = 1
Then in point-slope form, the equation of the tangent line is
... y = 1(x -1/2) +π/4
... y = x -1/2 +π/4
10. From a table of derivatives,
... arccos'(x) = -1/√(1 -x²)
Then, using the chain rule, ...
... arccos'(x²) = -2x/√(1 -x⁴)
For x=1/√2, the slope of the tangent is ...
... f'(1/√2) = -(2/√2)/√(1 -(1/√2)⁴) = (-√2)/√(3/4) = -2√(2/3)
... f'(1/√2) = (-2/3)√6
Then in point-slope form, the equation of the tangent line is
... y = (-2/3)(√6)(x -1/√2) +π/3
... y = (-2/3)(√6)x +(2/3)√3 +π/3
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Answer:
is the correct answer.
Step-by-step explanation:
Given:
To find:
if given term is written as following:
<u>Solution:</u>
We can see that the resulting expression does not contain anything under (square root) so we need to rationalize the denominator to remove the square root from denominator.
The rule to rationalize is:
Any term having square root term in the denominator, multiply and divide with the expression by changing the sign of square root term of the denominator.
Applying this rule to rationalize the given expression:
Comparing the above expression with:
A = 5, B = 6 (Not divisible by square of any prime)
C = -10
D = 2 (positive)
GCD of A, C and D is 1.
So,
