Determine the first derivative 1.1.
1.2.
The Correct Answer and Explanation is:
Here are the solutions for determining the first derivative of the given functions.
1.1. y = sec⁻¹((1+x²)/(1-x²))
Answer:
The first derivative is dy/dx = 2 / (1+x²).
Explanation:
To find the derivative of the function y = sec⁻¹((1+x²)/(1-x²)), we can use the chain rule. However, a more elegant approach is to first simplify the expression using a trigonometric substitution. The form of the argument (1+x²)/(1-x²) suggests a substitution involving the tangent function.
Let us make the substitution:
x = tan(θ)
This implies that θ = tan⁻¹(x).
Now, we substitute x = tan(θ) into the expression inside the inverse secant function:
Argument = (1 + tan²(θ)) / (1 – tan²(θ))
We use the fundamental trigonometric identity 1 + tan²(θ) = sec²(θ). The expression becomes:
Argument = sec²(θ) / (1 – tan²(θ))
Next, we can express this in terms of sine and cosine. Recall that sec(θ) = 1/cos(θ) and tan(θ) = sin(θ)/cos(θ).
Argument = (1/cos²(θ)) / (1 – sin²(θ)/cos²(θ))
Argument = (1/cos²(θ)) / ((cos²(θ) – sin²(θ))/cos²(θ))
Argument = 1 / (cos²(θ) – sin²(θ))
Using the double-angle identity for cosine, cos(2θ) = cos²(θ) – sin²(θ), the expression simplifies to:
Argument = 1 / cos(2θ) = sec(2θ)
Now, the original function can be rewritten in terms of θ:
y = sec⁻¹(sec(2θ))
The inverse secant and secant functions cancel each other out, leaving:
y = 2θ
Finally, we substitute θ = tan⁻¹(x) back into the equation to express y in terms of x:
y = 2tan⁻¹(x)
This simplified form of the function is much easier to differentiate. We now find the derivative dy/dx:
dy/dx = d/dx [2tan⁻¹(x)]
Using the standard derivative d/dx(tan⁻¹(x)) = 1 / (1+x²), we get:
dy/dx = 2 * (1 / (1+x²))
dy/dx = 2 / (1+x²)
1.2. y = ln[(1+tanh(x))/(1-tanh(x))]
Answer:
The first derivative is dy/dx = 2.
Explanation:
The function y = ln[(1+tanh(x))/(1-tanh(x))] can be differentiated using the chain rule and quotient rule. However, the calculation is significantly simplified by first using the properties of hyperbolic functions to simplify the argument of the natural logarithm.
We start with the argument of the logarithm:
Argument = (1 + tanh(x)) / (1 – tanh(x))
Recall the definitions of hyperbolic functions in terms of exponentials:
tanh(x) = (e^x – e^(-x)) / (e^x + e^(-x))
Substituting this into the argument can be cumbersome. A more direct simplification uses the identities derived from the exponential definitions. We know that:
cosh(x) + sinh(x) = e^x
cosh(x) – sinh(x) = e^(-x)
Let’s rewrite tanh(x) as sinh(x)/cosh(x) in our argument:
Argument = (1 + sinh(x)/cosh(x)) / (1 – sinh(x)/cosh(x))
Find a common denominator for the numerator and the denominator of the main fraction:
Argument = ((cosh(x) + sinh(x))/cosh(x)) / ((cosh(x) – sinh(x))/cosh(x))
The cosh(x) terms cancel out, leaving:
Argument = (cosh(x) + sinh(x)) / (cosh(x) – sinh(x))
Now, we use the exponential identities:
Argument = e^x / e^(-x) = e^(x – (-x)) = e^(2x)
The original function y simplifies dramatically to:
y = ln(e^(2x))
Using the property of logarithms that ln(e^a) = a, we get:
y = 2x
Now, we can find the first derivative of this simplified function, which is a straightforward application of the power rule:
dy/dx = d/dx(2x)
dy/dx = 2
This result shows that the original complex-looking function is actually a linear function with a constant slope of 2.
