Asked by Amber
Find an equation of the tangent line to the given curve at the specified point:
y=sqrt(x)/x+1, (4,2/5)
y=sqrt(x)/x+1, (4,2/5)
Answers
Answered by
Dr. Jane
You can find the slope of the tangent line by setting the first derivative equal to zero and solving for x.
rewrite sqrt x as x^1/2
This becomes the derivative of a quotient.
The denominator times the derivative of the numerator minus the numerator times the derivative of the denominator ALL OVER the denominator squared.
derivative of x^1/2 is (1/2)x^-1/2
derivative of x + 1 is 1
Can you put all of this together to find your slope?
Once you have the slope, find the equation of the line using the formula:
y = mx + b
you have x, m, y. Solve for b.
Then you will have the m and b that you need to substitute into y=mx + b
rewrite sqrt x as x^1/2
This becomes the derivative of a quotient.
The denominator times the derivative of the numerator minus the numerator times the derivative of the denominator ALL OVER the denominator squared.
derivative of x^1/2 is (1/2)x^-1/2
derivative of x + 1 is 1
Can you put all of this together to find your slope?
Once you have the slope, find the equation of the line using the formula:
y = mx + b
you have x, m, y. Solve for b.
Then you will have the m and b that you need to substitute into y=mx + b
Answered by
Amber
Is there anybody who can actually walk me through working out the problem please?
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