Can someone help me understand how to find the derivative of these two problems:
(t^2-1/t^2+2)^3=y and y=sin^32x?
I think with the first one, I could just foil it maybe and cancel but I don't know what to do with the ^3...and then the second one I haven't gotten the hang of how to eliminate those kinds of problems yet...:(
1)
y'=d/dt (uv^-1) where u= t^2-1 du=2dt
and v=(t^2+2)^3 dv= 3(t^2+2)^2 * (2t)=6t(t^2+2)^2
y'= v^-1 du -uv^-2 dv
y'=du/v -udv/v^2
then put in u,v du, dv and you have it.
I guess it's the format but I am little confused about what you did....
1)
use the quotient rule:
y = u/v
where u = t^2-1 and v = t^2+2
y' = (u'v - uv')/v^2
= [(2t)(t^2+2) - (t^2-1)(2t)]/(t^2+2)^2
= (2t^3+4t-2t^3+2t)/(t^2+2)^2
= 6t/(t^2+2)^2
2) use the chain rule:
y = u^3 where u = sin2x
y' = 3u^2 u'
= 3sin^2(2x) 2cos2x
= 6sin^2(2x) cos(2x)
or
3sin4x sin2x
Of course, I can help you with that! Finding derivatives involves applying certain rules and techniques. Let's go through each problem step by step:
Problem 1: (t^2-1/t^2+2)^3 = y
To find the derivative of this expression, we can apply the chain rule. The chain rule states that if we have a function raised to a power, we can differentiate it by first finding the derivative of the function and then multiplying it by the derivative of the power.
1. Start by rewriting the expression as y = (t^2-1)/(t^2+2) and then raise it to the power of 3: y^3 = [(t^2-1)/(t^2+2)]^3.
2. Next, find the derivative of the function that appears inside the parentheses, denoted by "d/dt(t^2-1)/(t^2+2)". To do this, we can use the quotient rule.
a. Apply the quotient rule: (d/dt(u))/(v) - u*(d/dt(v))/(v^2), where u = t^2 - 1 and v = t^2 + 2.
b. Determine the derivatives: (2t)(t^2 + 2) - (t^2 - 1)(2t)/(t^2 + 2)^2.
c. Simplify the expression: (2t^3 + 4t - 2t^3 + 1)/(t^2 + 2)^2 = (4t + 1)/(t^2 + 2)^2.
3. Now, differentiate the power of 3 using the power rule. The power rule states that if we have an expression raised to a constant power, we can multiply it by the constant and then subtract 1 from the power.
a. Apply the power rule: (3)(y^2)(dy/dt).
b. Substitute y with the expression obtained earlier: (3)[(t^2-1)/(t^2+2)]^2(dy/dt).
4. Finally, multiply the derivatives of the function and the power: (4t + 1)/(t^2 + 2)^2 * (3)[(t^2-1)/(t^2+2)]^2(dy/dt).
This is the derivative of y with respect to t.
Problem 2: y = sin^3(2x)
To find the derivative of this trigonometric function, we can use the chain rule as well.
1. Start by differentiating the outer function, which is sin^3(2x). To differentiate sin^3(2x), we can rewrite it as (sin(2x))^3.
2. Apply the chain rule by finding the derivative of the inner function and multiplying it by the derivative of the outer function.
a. Find the derivative of the inner function, which is 2.
b. Find the derivative of the outer function, which is 3(sin(2x))^2 * (cos(2x)), using the power rule and the chain rule.
3. Multiply the derivatives: dy/dx = 2 * 3(sin(2x))^2 * (cos(2x)).
This is the derivative of y with respect to x.
Remember, practice is key to mastering calculus. If you need further assistance or clarification, feel free to ask!