Find the derivative of xsecy+y^5=5x^3-7
x secy + y^5 = 5x^3 - 7
secy + x secy tany y' + 5y^4 y' = 15x^2
(x secy tany + 5y^4)y' = 15x^2 - secy
So,
y' =
15x^2 - secy
----------------------------
x secy tany + 5y^4
To find the derivative of the given equation, we need to differentiate both sides of the equation with respect to x.
The given equation is xsec(y) + y^5 = 5x^3 - 7.
Differentiate both sides of the equation with respect to x:
d/dx(xsec(y) + y^5) = d/dx(5x^3 - 7)
To differentiate, we use the chain rule and product rule on the left-hand side of the equation and the power rule on the right-hand side.
On the left-hand side:
d/dx(xsec(y)) = sec(y) * d/dx(x) + x * d/dx(sec(y))
Remember that d/dx(x) = 1, so the first term simplifies to sec(y).
To differentiate sec(y) with respect to x, we use the chain rule:
d/dx(sec(y)) = sec(y) * tan(y) * d/dx(y).
Combining the terms, we have:
sec(y) + x * sec(y) * tan(y) * d/dx(y) + d/dx(y^5) = 15x^2.
On the right-hand side:
d/dx(5x^3) = 15x^2 (using the power rule)
The constant term '-7' differentiates to zero.
Now, we need to find d/dx(y^5) and d/dx(y) separately.
d/dx(y^5) = 5y^4 * d/dx(y) (applying the power rule)
d/dx(y) is the derivative of y with respect to x. We treat y as a function of x and apply the chain rule:
d/dx(y) = d/dy(y) * d/dx(y) = 1 * d/dx(y) = d/dx(y)
Now, substitute these into the equation:
sec(y) + x * sec(y) * tan(y) * d/dx(y) + 5y^4 * d/dx(y) = 15x^2
To find d/dx(y), we need to isolate the term that contains it.
Rearrange the equation to solve for d/dx(y):
d/dx(y) = (15x^2 - sec(y))/ (x * sec(y) * tan(y) + 5y^4)
Thus, the derivative of the given equation with respect to x is (15x^2 - sec(y))/(x * sec(y) * tan(y) + 5y^4).