Derivative of Cosh

We now understand the derivative of cosh x. It is commonly represented as d/dx (cosh x) or (cosh x)', and its value is sinh x. The function cosh x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Hyperbolic Cosine Function: \( \cosh(x) = \frac{e^x + e^{-x}}{2} \). Hyperbolic Sine Function: \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).

The derivative of cosh x can be denoted as d/dx (cosh x) or (cosh x)'. The formula we use to differentiate cosh x is: d/dx (cosh x) = sinh x (or) (cosh x)' = sinh x The formula applies to all x.

We can derive the derivative of cosh x using proofs. To show this, we will use the hyperbolic identities along with the rules of differentiation. There are several methods we use to prove this, such as: By First Principle Using Chain Rule Using Product Rule We will now demonstrate that the differentiation of cosh x results in sinh x using the above-mentioned methods: By First Principle The derivative of cosh x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of cosh x using the first principle, we will consider f(x) = cosh x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = cosh x, we write f(x + h) = cosh (x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [cosh(x + h) - cosh x] / h = limₕ→₀ [ [(e^(x+h) + e^-(x+h))/2] - [(e^x + e^-x)/2] ] / h = limₕ→₀ [ (e^(x+h) - e^x)/2h + (e^-x - e^-(x+h))/2h ] = limₕ→₀ [ (e^x(e^h - 1)/2h) + (e^-x(1 - e^-h)/2h) ] Using limit formulas, limₕ→₀ (e^h - 1)/h = 1 and limₕ→₀ (1 - e^-h)/h = 1, f'(x) = e^x/2 + e^-x/2 = sinh x. Hence, proved. Using Chain Rule To prove the differentiation of cosh x using the chain rule, We use the formula: cosh x = (e^x + e^-x)/2 Let f(x) = e^x and g(x) = e^-x So, cosh x = (f(x) + g(x))/2 d/dx [cosh x] = 1/2 [d/dx(f(x)) + d/dx(g(x))] = 1/2 [e^x - e^-x] = sinh x. Using Product Rule We will now prove the derivative of cosh x using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, cosh x = (e^x + e^-x)/2 cosh x = 1/2 (e^x + e^-x) Given that, u = e^x and v = e^-x Using the product rule formula: d/dx [u + v] = u' + v' u' = d/dx (e^x) = e^x. (substitute u = e^x) v' = d/dx (e^-x) = -e^-x. (substitute v = e^-x) Using the product rule formula: d/dx (cosh x) = 1/2 (u' + v') = 1/2 [e^x - e^-x] = sinh x.

When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like cosh(x). For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues. For the nth Derivative of cosh(x), we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change (continuing for higher-order derivatives).

When x is 0, the derivative of cosh x = sinh(0), which is 0. When x approaches infinity, the derivative sinh(x) approaches infinity as well.

Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Hyperbolic Cosine Function: A hyperbolic function defined as cosh x = (e^x + e^-x)/2. Hyperbolic Sine Function: A hyperbolic function defined as sinh x = (e^x - e^-x)/2. First Derivative: It is the initial result of a function, which gives us the rate of change of a specific function. Chain Rule: A formula for computing the derivative of the composition of two or more functions.