Question

differentiate the function. y=((e^(5u))-e^(-5u))/((e^(5u))+e^(-5u))

Answer 1

Note that
y = ((e^(5u))-e^(-5u))/((e^(5u))+e^(-5u)) = tanh(5u)
so y' =sech^2(5u) = 4/((e^(5u))+e^(-5u))^2

Answer 2

To differentiate the given function, we will use the quotient rule.

Let's begin by simplifying the function:

y = ((e^(5u)) - e^(-5u))/((e^(5u)) + e^(-5u))

To differentiate the function y, we need to find the derivatives of the numerator and denominator.

For the numerator, let's first differentiate (e^(5u)):

d/dx(e^(5u)) = 5e^(5u)

Next, let's differentiate (-e^(-5u)):

d/dx(-e^(-5u)) = -(-5e^(-5u)) = 5e^(-5u)

Now let's find the derivative of the denominator:

d/dx((e^(5u)) + e^(-5u)) = d/dx(e^(5u)) + d/dx(e^(-5u))
= 5e^(5u) + 5e^(-5u)

Now we can apply the quotient rule:

dy/dx = (denominator * derivative of numerator - numerator * derivative of denominator) / (denominator)^2

dy/dx = ((e^(5u)) + e^(-5u))(5e^(5u) + 5e^(-5u)) - ((e^(5u)) - e^(-5u))(5e^(5u) + 5e^(-5u)) / ((e^(5u)) + e^(-5u))^2

Simplifying:

dy/dx = 10(e^(5u) + e^(-5))(e^(5u) + e^(-5u)) - 10(e^(5u) - e^(-5u))(e^(5u) + e^(-5u)) / ((e^(5u) + e^(-5u))^2

Finally, we can simplify this expression further if desired, but this is the first derivative of the given function.

Answer 3

You are correct, the given function can be simplified to y = tanh(5u), where tanh(u) represents the hyperbolic tangent function.

To find the derivative, we can differentiate tanh(5u) using the chain rule.

Let's denote f(u) = tanh(u) and g(u) = 5u.
Using the chain rule, the derivative of tanh(5u) with respect to u is:

dy/dx = (df/du)(dg/du) = sech^2(5u)(5) = 5sech^2(5u)

So the derivative of y = tanh(5u) is y' = 5sech^2(5u).

As an additional note, sech(x) is the hyperbolic secant function, which is defined as 1/cosh(x).