y = tan(sqrtx)
Find dy/dx.
So I found it and I got the answer as sec^2(sqrtx) but the answer is (sec^2(sqrtx))/2(sqrtx)...why?
----------------------------------
2) Find the line which passes through the point (0, 1/4) and is tangent to the curve y=x^3 at some point.
So I found the derivative which is 3x^2.
Let (a, a3) be the point of tangency.
3x^2 = (a3 - 1/4)/(a-0)
I'm not sure how to solve for a.
Can someone please help me out with these? I have an exam tomorrow evening and these questions are a lost cause.
y = tan u
dy/dx = sec^2 u du/ dx
here u = x^(1/2)
du/dx = sec^2 (x^.5) .5 x^-.5
=(1/2) (1/sqrt x) sec^2(sqrt x)
I agree with the authorities.
2) Find the line which passes through the point (0, 1/4) and is tangent to the curve y=x^3 at some point.
So I found the derivative which is 3x^2.
Let (a, a3) be the point of tangency.
3x^2 = (a3 - 1/4)/(a-0)
I'm not sure how to solve for a.
--------------------------------
sketching that I see no such line. Are you sure it is not through (1/4 , 0)
To find dy/dx for the function y = tan(sqrt(x)), you can use the chain rule.
Step 1: Differentiate the outer function:
The derivative of tan(x) is sec^2(x).
Step 2: Differentiate the inner function:
The derivative of sqrt(x) is (1/2)(1/sqrt(x)), which can be simplified to 1/(2sqrt(x)).
Step 3: Apply the chain rule:
dy/dx = (sec^2(sqrt(x))) * (1/(2sqrt(x)))
So the correct answer is (sec^2(sqrt(x))) / (2sqrt(x)). The additional factor of 1 / 2(sqrt(x)) arises from the chain rule.
Now, let's move on to the second question.
To find the equation of the line that passes through the point (0, 1/4) and is tangent to the curve y = x^3 at some point, we need to find the slope of the tangent line and then use the point-slope form of the equation of a line.
Step 1: Find the derivative of the curve:
The derivative of y = x^3 is dy/dx = 3x^2.
Step 2: Find the slope of the tangent line:
Let (a, a^3) be the point of tangency. The slope of the tangent line is equal to the derivative evaluated at that point, so we have:
3a^2 = (a^3 - 1/4)/(a - 0)
Step 3: Solve for 'a':
To solve this equation, we can multiply both sides by (a - 0) to get rid of the denominator:
3a^3 = a^3 - 1/4
Combine like terms:
2a^3 = -1/4
Divide both sides by 2 to isolate 'a':
a^3 = -1/8
Take the cube root of both sides:
a = -1/2
Step 4: Write the equation of the tangent line using the point-slope form:
Using the point (0, 1/4) and the slope found above, we have:
y - 1/4 = 3*(-1/2)^2 * (x - 0)
Simplifying the equation gives:
y - 1/4 = 3/4 * x
Multiply through by 4 to get rid of the fraction:
4y - 1 = 3x
So the equation of the line that passes through (0, 1/4) and is tangent to the curve y = x^3 at some point is 4y - 1 = 3x.
I hope this helps! Good luck with your exam. Let me know if you have any further questions.