Can someone help me find the derivatives of these questions?
For these can you only need to tell me the answer so i can see if i have the same thing.
y=cos(-4x)
y = sinx^2
y = xcosx
I need help with these 6 please.
y = xcos(1/x)
y = (1+sinx)/(1-sin^(2)x)
y = sin^2x/cosx
y = sqrt(sin(sqrtx))
y = 3sqrt(xcosx)
y = 1/(sin(x-sinx)
y=cos(-4x)
y' = -4(-sin(-4x)
= 4sin(-4x)
y = sinx^2 or (sinx)^2
y' = 2sinxcosx or sin 2x
y = xcosx
y' = x(-sinx) + (1)cosx
= cosx - xsinx
for y = xcos(1/x)
I would use the product rule
y' = x(-sin(1/x))(-1/x^2) + (1)cos(1/x)
for y = (1+sinx)/(1-sin^(2)x) I would reduce it to
y = (1+sinx)/((1+sinx)(1-sinx))
= 1/(1-sinx)
= (1-sinx)^-1
now y' = -1(1-sinx)-2?sup>(-cosx)
why don't you show me some of the steps you have sofar for the others?
last line should have come out as ...
y = -1(1-sinx)-2(-cosx)
Sure! I can help you find the derivatives of these functions. Here's how you can find the derivatives step-by-step:
1. y = cos(-4x)
To find the derivative of y with respect to x, you can use the chain rule. The derivative of cos(u) with respect to u is -sin(u). So, we can calculate the derivative as follows:
dy/dx = d(cos(-4x))/dx = -sin(-4x) = sin(4x).
2. y = sin(x^2)
To find the derivative of y with respect to x, you can use the chain rule. The derivative of sin(u) with respect to u is cos(u). So, we can calculate the derivative as follows:
dy/dx = d(sin(x^2))/dx = cos(x^2) * d(x^2)/dx = cos(x^2) * 2x = 2xcos(x^2).
3. y = xcosx
For this function, you can use the product rule to find the derivative. The product rule states that if y = f(x)g(x), then the derivative of y with respect to x is given by:
dy/dx = f'(x)g(x) + f(x)g'(x).
Using this rule, we have:
dy/dx = 1 * cos(x) + x * (-sin(x)) = cos(x) - xsin(x).
4. y = xcos(1/x)
To find the derivative of y with respect to x, you can again use the chain rule. The derivative of cos(u) with respect to u is -sin(u). Additionally, when differentiating 1/x, you can use the power rule for derivatives, which states that the derivative of x^n is n * x^(n-1) (where n is any constant). With these rules, we can calculate the derivative as follows:
dy/dx = d(xcos(1/x))/dx = -sin(1/x) * d(1/x)/dx + cos(1/x) * d(x)/dx = -sin(1/x) * (-1/x^2) + cos(1/x) = sin(1/x)/x^2 + cos(1/x).
5. y = (1+sinx)/(1-sin^2x)
To find the derivative of this function, we can simplify it first. Since sin^2x = (sinx)^2, we can rewrite the function as y = (1+sinx)/(1-(sinx)^2). This is equivalent to y = (1+sinx)/(cos^2x), since 1 - (sinx)^2 = cos^2x. Now, to find the derivative, we can use the quotient rule:
dy/dx = [((1+sinx) * d(cos^2x)/dx) - (cos^2x * d(1+sinx)/dx)] / (cos^2x)^2
Simplifying further, we have:
dy/dx = [((1+sinx) * (-2sinx * cosx)) - (cos^2x * cosx)] / cos^4x
= [-2sinx * cosx - cos^3x * cosx + sinx] / cos^4x
= [-2sinx * cosx - cos^4x * cosx + sinx] / cos^4x.
6. y = 1/(sin(x-sinx))
To find the derivative of y with respect to x, we can use the chain rule. The derivative of sin(u) with respect to u is cos(u). With this rule, we can calculate the derivative as follows:
dy/dx = d(1/(sin(x-sinx)))/dx = -1/(sin^2(x-sinx)) * d(sin(x-sinx))/dx
= -1/(sin^2(x-sinx)) * cos(x-sinx) * d(x-sinx)/dx
= -cos(x-sinx) / (sin^2(x-sinx)).
Remember to always double-check the solutions by comparing them to confirm if they match the ones you get. I hope this helps!