Find dy/dx, if y= 5e^x/cosx
y = (5e^x)/(cos x)
Since 1/cos(x) = sec(x), we can also rewrite this as
(5e^x)(sec(x))
Since two terms which are both functions of x are multiplied, we use the chain rule. Thus it becomes:
(5e^x)(sec(x)) + (5e^x)(tan(x) * sec(x))
Simplifying,
dy/dx = (5e^x)(sec(x))(1 + tan(x))
Hope this helps~ :3
Let f(x) = e^x
So f'(x) = e^x
Let g(x) = cos(x)
So g'(x) = -sin(x)
Use:
(f/g)' = ((f'×g)-(f×g'))/g^2
To find the derivative dy/dx of the given function, y = 5e^x/cos(x), we need to use the quotient rule and the chain rule.
The quotient rule states that if we have a function in the form f(x) = g(x) / h(x), then the derivative is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Let's begin by finding the derivative of the numerator and denominator separately.
1. For g(x) = 5e^x:
The derivative of e^x with respect to x is simply e^x. Since there is a coefficient of 5, we multiply it with the derivative of e^x. Therefore,
g'(x) = 5e^x
2. For h(x) = cos(x):
The derivative of cos(x) with respect to x is -sin(x). Therefore,
h'(x) = -sin(x)
Now, let's substitute these values into the quotient rule:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
= (5e^x * cos(x) - 5e^x * (-sin(x))) / (cos(x))^2
= (5e^x * cos(x) + 5e^x * sin(x)) / (cos(x))^2
Therefore, the derivative dy/dx is given by:
dy/dx = (5e^x * cos(x) + 5e^x * sin(x)) / (cos(x))^2
To find dy/dx when y = 5e^x/cosx, we can use the quotient rule. The quotient rule states that if we have a function u(x) = f(x)/g(x), where f(x) and g(x) are both differentiable functions, then the derivative of u(x) with respect to x, denoted as du/dx, is given by the formula:
du/dx = (g(x)*d(f(x))/dx - f(x)*d(g(x))/dx) / (g(x))^2
Now, let's apply the quotient rule to find dy/dx for y = 5e^x/cosx.
First, we need to find the derivatives of the numerator and denominator separately.
The derivative of f(x) = 5e^x is d(f(x))/dx = 5e^x.
The derivative of g(x) = cosx is d(g(x))/dx = -sinx.
Now, plug these derivatives into the quotient rule formula:
dy/dx = [(cosx)(5e^x) - (5e^x)(-sinx)] / (cosx)^2
Simplifying further:
dy/dx = (5e^xcosx + 5e^xsinx) / (cosx)^2
Therefore, dy/dx = (5e^xcosx + 5e^xsinx) / (cosx)^2 is the derivative of y with respect to x.