## d/dx (tanh x)

### To find the derivative of the hyperbolic tangent function, tanh(x), with respect to x, we can use the chain rule

To find the derivative of the hyperbolic tangent function, tanh(x), with respect to x, we can use the chain rule. Let’s break down the process step by step:

Step 1: Recall the definition of the hyperbolic tangent function, tanh(x), which is given by:

tanh(x) = (e^(2x) – 1) / (e^(2x) + 1)

Step 2: Rewrite tanh(x) using an exponential form:

tanh(x) = (e^(2x) – 1) / (e^(2x) + 1) = (e^(2x) – 1) * (e^(-2x) / (e^(2x) + 1))

Step 3: Simplify the expression:

tanh(x) = (e^(2x) – e^(-2x)) / (e^(2x) + 1)

Step 4: Now, let’s find the derivative of tanh(x) using the quotient rule. The quotient rule states that if we have a function of the form f(x) = g(x) / h(x), then its derivative is given by:

f'(x) = (g'(x) * h(x) – g(x) * h'(x)) / (h(x))^2

Applying the quotient rule, let:

g(x) = e^(2x) – e^(-2x)

h(x) = e^(2x) + 1

g'(x) = (d/dx)(e^(2x) – e^(-2x))

= 2e^(2x) + 2e^(-2x) (Using the derivative of the exponential function)

= 2(e^(2x) + e^(-2x))

h'(x) = (d/dx)(e^(2x) + 1)

= 2e^(2x) (Using the derivative of the exponential function)

Now, using the quotient rule formula, we have:

(tanh(x))’ = (g'(x) * h(x) – g(x) * h'(x)) / (h(x))^2

= (2(e^(2x) + e^(-2x)) * (e^(2x) + 1) – (e^(2x) – e^(-2x)) * 2e^(2x)) / (e^(2x) + 1)^2

= (2(e^(4x) + 2 + 1 – e^(2x)) – (e^(4x) – 1)) / (e^(2x) + 1)^2

= (e^(4x) + 2 – 2e^(2x) – e^(4x) + 1) / (e^(2x) + 1)^2

= (3 – 2e^(2x)) / (e^(2x) + 1)^2

Therefore, the derivative of the hyperbolic tangent function, d/dx (tanh(x)), is:

d/dx (tanh(x)) = (3 – 2e^(2x)) / (e^(2x) + 1)^2

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