Posted by Samantha on Tuesday, November 4, 2008 at 8:01pm.
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)
Related Questions
Calculus - lim(x->infinity) sqrtx sin(1/sqrtx) Someone please start me ...
Calculus - I don't know how to do the integral of e^(lnx^2)dx and the ...
math - Please help me simplify: 4sqrtx^12/24sqrtx^2 *Is this correct: 2sqrt4/12x...
Integration - Intergrate ¡ì sec^3(x) dx could anybody please check ...
calculus - find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx ...
Business Cal - Suppose a publishing company has found that the marginal cost at ...
calculus (check my work please) - Not sure if it is right, I have check with the...
finding c: calc - let f(x)=square root x. if the rate of change of at x=c is ...
Calculus 1A - I am trying to find: dy/dx for y = sec(tan x) I have the answer, ...
Calculus - Find the partial derivative for x. f(x,y)=1/SQRTx^2+y^2
For Further Reading
