Hi, I have few problems that I really need help solving. I think I start out right but get messed up along the way and get the wrong answer...Any help is appreciated! thank you so much!
find derivative of:
1) y=arctan 2^x
2) arctan(xy)= 1+x^2y
Also...
3) find the equation of the tangent line to the curve y= 3arccos(x/2) at the point (1,ð)
2) Differentiate arctan(xy)= 1+x^2y
to obtain dy/dx. Use implicit differentiation. Take the derivateve of both sides of the equation with respect to x, remembering that y is a function of x.
The derivative of arctan u is 1/(1+u^2).
1/[1 + (xy)^2]*[x*dy/dx + y]
= x^2*dy/dx + 2xy
y/[1 + (xy)^2] - 2 xy
= dy/dx*{-x/[1 + (xy)^2]+ x^2}
dy/dx = {y/[1 + (xy)^2] - 2 xy}/{-x/[1 + (xy)^2]+ x^2}
Let 2^x = u(x); du/dx = 2^x *ln2
y = arctan u(x)
dy/dx = dy/du*du/dx
= [1/(1+u^2)]*2^x*ln2
= [1/(1+2^(2x))]* 2^x * ln2
Sure, I'd be happy to help you with these problems! Let's go through each of them step by step and find the solutions together.
1) To find the derivative of y = arctan(2^x), we can use the chain rule. The chain rule states that if we have a composite function like f(g(x)), then the derivative of f(g(x)) with respect to x is given by f'(g(x)) * g'(x).
In this case, let's define u = 2^x. Then y can be written as y = arctan(u). Now, let's find the derivatives of u with respect to x and y with respect to u.
The derivative of u with respect to x can be found using the power rule: du/dx = (ln(2)) * (2^x).
The derivative of y with respect to u can be found using the fact that the derivative of arctan(u) is 1/(1+u^2): dy/du = 1/(1+u^2).
Finally, applying the chain rule, we can find the derivative of y with respect to x as:
dy/dx = (dy/du) * (du/dx)
= (1/(1+u^2)) * (ln(2)) * (2^x)
So, the derivative of y = arctan(2^x) is dy/dx = (1/(1+u^2)) * (ln(2)) * (2^x) where u = 2^x.
2) To find the derivative of arctan(xy) = 1+x^2y, we can again use the chain rule.
Let's define u = xy. Then the equation can be rewritten as arctan(u) = 1+x^2y. Now, let's find the derivatives of u with respect to x and y with respect to u.
The derivative of u with respect to x is simply y.
The derivative of y with respect to u can be found using the fact that the derivative of arctan(u) is 1/(1+u^2): dy/du = 1/(1+u^2).
Finally, applying the chain rule, we can find the derivative of y with respect to x as:
dy/dx = (dy/du) * (du/dx)
= (1/(1+u^2)) * y
Substituting u = xy back in, we get:
dy/dx = (1/(1+(xy)^2)) * y
So, the derivative of arctan(xy) = 1+x^2y is dy/dx = (1/(1+(xy)^2)) * y.
3) To find the equation of the tangent line to the curve y = 3arccos(x/2) at the point (1, π), we need to find the slope of the tangent line at that point.
First, let's find the derivative of y with respect to x using the chain rule. The derivative of arccos(x/2) can be found using the chain rule and gives us -1/(√(1-(x/2)^2)). So, the derivative of y = 3arccos(x/2) is dy/dx = 3 * (-1/(√(1-(x/2)^2))).
Now, let's find the slope of the tangent line by substituting x = 1 into the derivative:
m = dy/dx (at x=1)
= 3 * (-1/(√(1-(1/2)^2)))
= 3 * (-1/(√(1-1/4)))
= 3 * (-1/(√(3/4)))
= 3 * (-1/(√3/2))
= -2√3
So, the slope of the tangent line at x = 1 is -2√3.
Now we can find the equation of the tangent line using the point-slope form of a line: y - y1 = m(x - x1), where (x1, y1) is the point (1, π).
Plugging in the values, we have:
y - π = -2√3(x - 1)
And that's the equation of the tangent line to the curve y = 3arccos(x/2) at the point (1, π).