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Derivative of Hyperbolic Functions
Before getting into the details of the derivative of hyperbolic functions, let us recall the concept of the hyperbolic functions. Hyperbolic functions are functions in calculus that are expressed as combinations of the exponential functions e^{x} and e^{x}. We have six main hyperbolic functions given by, sinhx, coshx, tanhx, sechx, cothx, and cschx. The derivative of hyperbolic functions is calculated using the derivatives of exponential functions formula and other hyperbolic functions formulas and identities.
In this article, we will evaluate the derivatives of hyperbolic functions using different hyperbolic trig identities and derive their formulas. We will also explore the graphs of the derivative of hyperbolic functions and solve examples and find derivatives of functions using these derivatives for a better understanding of the concept.
What is Derivative of Hyperbolic Functions?
The derivative of hyperbolic functions gives the rate of change in the hyperbolic functions as differentiation of a function determines the rate of change in function with respect to the variable. We can evaluate these derivatives using the derivative of exponential functions e^{x} and e^{x} along with other hyperbolic functions formulas and identities. We have six main hyperbolic functions namely,
 Sinhx
 Coshx
 Tanhx
 Cothx
 Sechx
 Cschx
The derivative of hyperbolic functions is used in describing the shape of electrical wires hanging freely between two poles. They are also used to describe any freely hanging cable between two ends. Among other applications, the derivative of hyperbolic functions is used to describe the formation of satellite rings and planets. In the next section, we will explore the formulas of the derivatives of hyperbolic functions.
Derivative of Hyperbolic Functions Formula
Now, we will go through the formulas of the derivatives of hyperbolic functions. The hyperbolic functions are combinations of exponential functions e^{x} and e^{x}. Given below are the formulas for the derivative of hyperbolic functions:
 Derivative of Hyperbolic Sine Function: d(sinhx)/dx = coshx
 Derivative of Hyperbolic Cosine Function: d(coshx)/dx = sinhx
 Derivative of Hyperbolic Tangent Function: d(tanhx)/dx = sech^{2}x
 Derivative of Hyperbolic Cotangent Function: d(cothx)/dx = csch^{2}x (x ≠ 0)
 Derivative of Hyperbolic Secant Function: d(sechx)/dx = sechx tanhx
 Derivative of Hyperbolic Cosecant Function: d(cschx)/dx = cschx cothx (x ≠ 0)
Let us now prove these derivatives using different mathematical formulas and identities.
Derivatives of Hyperbolic Functions Proof
Now that we know the formulas for the derivatives of hyperbolic functions, let us now prove them using various formulas and identities of hyperbolic functions. We will use the following formulas to prove the derivative of hyperbolic functions:
 Derivative of e to the power x: d(e^{x})/dx = e^{x}
 Derivative of e to the power negative x: d(e^{x})/dx = e^{x}
Derivative of Sinhx
We know that the formula for sinhx is given by, sinhx = (e^{x}  e^{x})/2. To find the derivative of hyperbolic function sinhx, we will write as a combination of exponential function and differentiate it using the quotient rule of differentiation. Also, we know that we can write the hyperbolic function cosh x as cosh x = (e^{x} + e^{x})/2. So, using these formulas, we have
d(sinhx)/dx = d[(e^{x}  e^{x})/2] / dx
= [ (e^{x}  e^{x})' 2  (e^{x}  e^{x}) (2)' ] / 2^{2}
= [e^{x}  (e^{x})] 2 / 2^{2}  [Using d(e^{x})/dx = e^{x} and d(e^{x})/dx = e^{x}]
= (e^{x} + e^{x})/2
= cosh x
Therefore, the derivative of sinh x is equal to cosh x.
Derivative of Coshx
To prove the derivative of coshx, we will use the following formulas:
 sinhx = (e^{x}  e^{x})/2
 cosh x = (e^{x} + e^{x})/2
 d(e^{x})/dx = e^{x}
 d(e^{x})/dx = e^{x}
Using the above formulas, we have
d(coshx)/dx = d[(e^{x} + e^{x})/2] / dx
= d(e^{x}/2) / dx + d(e^{x}/2) / dx
= e^{x}/2  e^{x}/2
= (e^{x}  e^{x})/2
= sinhx
Hence, we have proved that the derivative of coshx is equal to sinhx.
Derivative of Tanhx
Using hyperbolic functions formulas, we know that tanhx can be written as the ratio of sinhx and coshx. So, we will use the quotient rule and the following formulas to find the derivative of tanhx:
 tanhx = sinhx / coshx
 d(sinhx)/dx = coshx
 d(coshx)/dx = sinhx
 cosh^{2}x  sinh^{2}x = 1
 1/coshx = sechx
Using the above formulas, we have
d(tanhx)/dx = d(sinhx / coshx)/dx
= [(sinhx)^{2} coshx  (coshx)' sinhx] / cosh^{2}x  [Using quotient rule of derivatives]
= (cosh^{2}x  sinh^{2}x) / cosh^{2}x
= 1 / cosh^{2}x
= sech^{2}x
Hence, the derivative of hyperbolic function tanhx is equal to sech^{2}x.
Derivative of Cothx
Just like we derived the derivative of tanhx, we will evaluate the derivative of hyperbolic function cothx using the quotient rule. Also, we can express cothx as the ratio of coshx and sinhx. We will use the following formulas to calculate the derivative of tanhx:
 cothx = coshx/sinhx
 d(sinhx)/dx = coshx
 d(coshx)/dx = sinhx
 cosh^{2}x  sinh^{2}x = 1
 1/sinhx = cschx
So, we have
d(cothx)/dx = d(coshx/sinhx)/dx
= [(coshx)' sinhx  (sinhx)' coshx] / sinh^{2}x
= (sinh^{2}x  cosh^{2}x) / sinh^{2}x
= (cosh^{2}x  sinh^{2}x) / sinh^{2}x
= 1 / sinh^{2}x
=  csch^{2}x
Therefore, the derivative of cothx is equal to  csch^{2}x.
Derivative of Sechx
In this section, we will derive the formula for the derivative of sechx using the quotient rule. We will use the formula for the derivative of coshx along with other formulas given by,
 d(coshx)/dx = sinhx
 sechx = 1/coshx
 tanhx = sinhx/coshx
Using the above formulas, we have
d(sechx)/dx = d(1/coshx)/dx
= [(1)' coshx  (coshx)' (1)] / cosh^{2}x
= (0 × coshx  sinhx) / cosh^{2}x
= sinhx / cosh^{2}x
=  (sinhx / coshx) × (1/coshx)
=  tanhx sechx
Hence, the derivative of hyperbolic function sechx is equal to  tanhx sechx.
Derivative of Cschx
To find the derivative of cschx, we will use a similar method as we used to find the derivative of sechx. We will use the following formulas to find the derivative of cschx:
 d(sinhx)/dx = coshx
 1/sinhx = cschx
 cothx = coshx/sinhx
So, we have
d(cschx)/dx = d(1/sinhx)/dx
= [(1)' sinhx  (sinhx)' (1)] / sinh^{2}x
= (0 × sinhx  coshx) / sinh^{2}x
= coshx / sinh^{2}x
=  (coshx / sinhx) × (1/sinhx)
=  cothx cschx (x ≠ 0)
Hence, we have proved that the derivative of cschx is equal to  cothx cschx.
Derivative of Inverse Hyperbolic Functions
Now that we have derived the derivative of hyperbolic functions, we will derive the formulas of the derivatives of inverse hyperbolic functions. We can find the derivatives of inverse hyperbolic functions using the implicit differentiation method. We have six main inverse hyperbolic functions, given by arcsinhx, arccoshx, arctanhx, arccothx, arcsechx, and arccschx. Their derivatives are given by:
 Derivative of arcsinhx: d(arcsinhx)/dx = 1/√(x^{2} + 1), ∞ < x < ∞
 Derivative of arccoshx: d(arccoshx)/dx = 1/√(x^{2}  1), x > 1
 Derivative of arctanhx: d(arctanhx)/dx = 1/(1  x^{2}), x < 1
 Derivative of arccothx: d(arccothx)/dx = 1/(1  x^{2}), x > 1
 Derivative of arcsechx: d(arcsechx)/dx = 1/x√(1  x^{2}), 0 < x < 1
 Derivative of arccschx: d(arccschx)/dx = 1/x√(1 + x^{2}), x ≠ 0
Now, let use derive the above formulas of derivatives of inverse hyperbolic functions using implicit differentiation method.
Derivative of Arcsinhx
Assume arcsinhx = y, then we have x = sinh y. Now, differentiating both sides of x = sinh y with respect to x, we have:
dx/dx = d(sinh y)/dx
⇒ 1 = cosh y × dy/dx  [Because derivative of sinh y is cosh y]
⇒ dy/dx = 1/coshy
= 1/√(1 + sinh^{2}y)  [Because cosh^{2}A  sinh^{2}A = 1 which implies coshA = √(1 + sinh^{2}A)]
= 1/√(1 + x^{2})  [Because x = sinh y]
⇒ d(arcsinhx)/dx = 1/√(1 + x^{2})
Derivative of Arccoshx
To find the derivative of arccoshx, we assume arccoshx = y. This implies we have x = cosh y. Now, differentiating both sides of x = cosh y, we have
dx/dx = d(cosh y)/dx
⇒ 1 = sinh y × dy/dx  [Because derivative of cosh y is sinh y]
⇒ dy/dx = 1/sinhy
= 1/√(cosh^{2}y  1)  [Because cosh^{2}A  sinh^{2}A = 1 which implies sinhA = √(cosh^{2}A  1)]
= 1/√(x^{2 } 1)  [Because x = cosh y]
⇒ d(arccoshx)/dx = 1/√(x^{2 } 1), x > 1
Derivative of Arctanhx
Next, we will calculate the derivative of tanhx. Assume arctanhx = y, then we have x = tanh y. Now, differentiating both sides of x = tanh y with respect to x, we have
dx/dx = d(tanh y)/dx
⇒ 1 = sech^{2}y × dy/dx  [Because derivative of tanh y is sech^{2}y]
⇒ dy/dx = 1/sech^{2}y
= 1/(1  tanh^{2}y)  [Using hyperbolic trig identity 1  tanh^{2}A = sech^{2}A]
= 1/(1  x^{2})  [Because x = tanhy]
⇒ d(arctanhx)/dx = 1/(1  x^{2}), x < 1
Derivative of Arccothx
We will find the derivative of arccothx using a similar way as we did for the derivative of arctanhx. Assume arccothx = y, then we have x = coth y. Now, differentiating both sides of x = coth y with respect to x, we have
dx/dx = d(coth y)/dx
⇒ 1 = csch^{2}y × dy/dx  [Because derivative of coth y is csch^{2}y]
⇒ dy/dx = 1/csch^{2}y
= 1/(coth^{2}y  1)  [Using hyperbolic trig identity coth^{2}A  1 = csch^{2}A]
= 1/(x^{2}  1)  [Because x = cothy]
⇒ d(arccothx)/dx = 1/(x^{2}  1) , x > 1
Derivative of Arcsechx
To find the derivative of arcsechx, we will use the formula for the derivative of sechx. Assume arcsechx = y, this implies we have x = sech y. Now, differentiating both sides of x = sech y with respect to x, we have
dx/dx = d(sech y)/dx
⇒ 1 = sech y tanh y × dy/dx  [Because derivative of sech y is sech y tanh y]
⇒ dy/dx = 1/sech y tanh y
= 1/sech y √(1  sech^{2}y)  [Using hyperbolic trig identity 1  tanh^{2}A = sech^{2}A which implies tanh A = √(1  sech^{2}A)]
= 1/x √(1  x^{2})
⇒ d(arcsechx)/dx = 1/x √(1  x^{2}) , 0 < x < 1
Derivative of Arccschx
To find the derivative of arccschx, we will use the formula for the derivative of cschx. Assume arccschx = y, this implies we have x = csch y. Now, differentiating both sides of x = csch y with respect to x, we have
dx/dx = d(csch y)/dx
⇒ 1 = csch y coth y × dy/dx  [Because derivative of sech y is csch y coth y]
⇒ dy/dx = 1/csch y coth y
= 1/csch y √(csch^{2}y + 1) [Using hyperbolic trig identity coth^{2}A  1 = csch^{2}A which implies coth A = ±√(csch^{2}A + 1)]
= 1/x √(x^{2} + 1)
⇒ d(arccschx)/dx = 1/x √(x^{2} + 1) , x ≠ 0
Derivatives of Hyperbolic Functions and Inverse Hyperbolic Functions Table
In the above sections, we have derived the formulas for the derivatives of hyperbolic functions and inverse hyperbolic functions. Let us now summarize all the derivatives in a table below along with their domains (restrictions):
Function  Derivative  Domain 

sinhx  coshx  ∞ < x < ∞ 
coshx  sinhx  ∞ < x < ∞ 
tanhx  sech^{2}x  ∞ < x < ∞ 
cothx  csch^{2}x  x ≠ 0 
sechx  sechx tanhx  ∞ < x < ∞ 
cschx  cschx cothx  x ≠ 0 
arcsinhx  1/√(x^{2} + 1)  ∞ < x < ∞ 
arccoshx  1/√(x^{2}  1)  x > 1 
arctanhx  1/(1  x^{2})  x < 1 
arccothx  1/(1  x^{2})  x > 1 
arcsechx  1/x√(1  x^{2})  0 < x < 1 
arccschx  1/x√(1  x^{2})  x ≠ 0 
Important Notes on Derivative of Hyperbolic Functions
 d(sinhx)/dx = coshx
 d(coshx)/dx = sinhx
 d(tanhx)/dx = sech^{2}x
 d(cothx)/dx = csch^{2}x (x ≠ 0)
 d(sechx)/dx = sechx tanhx
 d(cschx)/dx = cschx cothx (x ≠ 0)
☛ Related Topics:
Derivatives of Hyperbolic Functions Examples

Example 1: Find the derivative of hyperbolic function f(x) = sinhx + 2coshx
Solution: To find the derivative of f(x) = sinhx + 2coshx, we will use the following formulas:
 d(sinhx)/dx = coshx
 d(coshx)/dx = sinhx
We have
d(sinhx + 2coshx)/dx = d(sinhx)/dx + d(2coshx)/dx
= coshx + 2sinhx
Answer: Derivative of sinhx + 2coshx is equal to coshx + 2sinhx.

Example 2: Calculate the derivative of f(x) = 2x^{5}tanhx.
Solution: To find the derivative of f(x) = 2x^{5}tanhx, we will use the product rule, power rule and formula for the derivative of hyperbolic function tanhx.
d(2x^{5}tanhx)/dx = 2 [ (x^{5})' tanhx + x^{5} (tanhx)' ]
= 2[5x^{4} tanhx + x^{5} sech^{2}x]
= 2x^{4} (5 tanhx + x sech^{2}x)
Answer: The derivative of f(x) = 2x^{5}tanhx is 2x^{4} (5 tanhx + x sech^{2}x).

Example 3: Find the derivative of sinh x / (x + 1).
Solution: We will use the quotient rule to find this derivative. We know that derivative of hyperbolic function sinhx is equal to coshx. So, we have
d[sinh x / (x + 1)] / dx = [(sinh x)' (x + 1)  sinhx (x + 1)'] / (x + 1)^{2}
= [coshx (x + 1)  sinhx] / (x + 1)^{2}
Answer: Derivative of sinh x / (x + 1) is equal to [coshx (x + 1)  sinhx] / (x + 1)^{2}
FAQs on Derivative of Hyperbolic Functions
What are Derivatives of Hyperbolic Functions?
The derivative of hyperbolic functions gives the rate of change in the hyperbolic functions as differentiation of a function determines the rate of change in function with respect to the variable. We have six main hyperbolic functions given by, sinhx, coshx, tanhx, sechx, cothx, and cschx.
What is the Derivative of Hyperbolic Functions Formula?
Given below are the formulas for the derivative of hyperbolic functions:
 Derivative of Sinhx: d(sinhx)/dx = coshx
 Derivative of Coshx: d(coshx)/dx = sinhx
 Derivative of Tanhx: d(tanhx)/dx = sech^{2}x
 Derivative of Cothx: d(cothx)/dx = csch^{2}x (x ≠ 0)
 Derivative of Sechx: d(sechx)/dx = sechx tanhx
 Derivative of Cschx: d(cschx)/dx = cschx cothx (x ≠ 0)
How to Prove Derivatives of Hyperbolic Functions?
We can prove the derivative of hyperbolic functions by using the derivative of exponential function along with other hyperbolic formulas and identities. We know that hyperbolic functions are expressed as combinations of e^{x} and e^{x}.
What are Derivatives of Inverse Hyperbolic Functions?
The derivatives of inverse hyperbolic functions are given by:
 Derivative of arcsinhx: d(arcsinhx)/dx = 1/√(x^{2} + 1), ∞ < x < ∞
 Derivative of arccoshx: d(arccoshx)/dx = 1/√(x^{2}  1), x > 1
 Derivative of arctanhx: d(arctanhx)/dx = 1/(1  x^{2}), x < 1
 Derivative of arccothx: d(arccothx)/dx = 1/(1  x^{2}), x > 1
 Derivative of arcsechx: d(arcsechx)/dx = 1/x√(1  x^{2}), 0 < x < 1
 Derivative of arccschx: d(arccschx)/dx = 1/x√(1 + x^{2}), x ≠ 0
How Do You Take the Derivative of Hyperbolic Function Sinx?
We can find the derivative of sinhx by expressing it as d(sinhx)/dx = (e^{x}  e^{x})/2. So, we have d(sinhx)/dx = d[(e^{x}  e^{x})/2] / dx = (e^{x} + e^{x})/2 = cosh x.
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