y = tan^-1(sqrt(5x^2-1)) find dy/dx
just use the chain rule
1/(1+(√(5x^2-1))^2) * 5x/√(5x^2-1)
= 5x / √(5x^2-1)(1+5x^2-1)
= 1/x√(5x^2-1)
You can also do it by implicit differentiation
tan y = √(5x^2-1)
sec^2 y y' = 5x/√(5x^2-1)
since tan y = √(5x^2-1), sec^2y = 1+tan^2 = 5x^2
y' = 5x/(5x^2√(5x^2-1))
= 1/x√(5x^2-1)
To find the derivative of the given function y = tan^(-1)(√(5x^2-1)), we can use the chain rule and the derivatives of the inverse tangent and square root functions.
Step 1: Rewrite the function
First, let's rewrite the original equation using the chain rule and the composite function notation to make it easier to take the derivative.
Let u = √(5x^2-1), so now we have y = tan^(-1)(u).
Step 2: Find dy/du
The derivative of the inverse tangent function tan^(-1)(u) is given by dy/du = 1/(1 + u^2).
So, dy/du = 1/(1 + (√(5x^2-1))^2).
Step 3: Find du/dx
To find du/dx, we need to differentiate u with respect to x. Since u = √(5x^2-1), we can use the chain rule once again.
d(u)/d(x) = d(√(5x^2-1))/d(x).
Applying the chain rule, we get d(u)/d(x) = (1/2)*(1/√(5x^2-1))*(d(5x^2-1)/d(x)).
Simplifying further, we have d(u)/d(x) = (1/2)*(1/√(5x^2-1))*(10x).
Step 4: Find dy/dx
To find the derivative dy/dx, we can use the chain rule again.
dy/dx = (dy/du) * (du/dx).
Substituting the values obtained in Steps 2 and 3, we have
dy/dx = (1/(1 + (√(5x^2-1))^2))*(1/2)*(1/√(5x^2-1))*(10x).
Simplifying the expression further, we get
dy/dx = (10x)/(2(1 + (√(5x^2-1))^2)√(5x^2-1)).
Hence, the derivative dy/dx of the function y = tan^(-1)(√(5x^2-1)) is given by (10x)/(2(1 + (√(5x^2-1))^2)√(5x^2-1)).