How do you find the derivative of the function: #tan(arcsin x)#?

Question

Evelyn Agard · Accepted Answer

#Sec^2(arcsin x)/sqrt(1-x^2)#

or #Sec^2(sin^-1 x)/sqrt(1-x^2)# to find the derivative of the function #tan(arcsin x)#, use the following formulas for differentiation #d/dx(tan u) = sec^2 u * d/dx(u)# and #d/dx(arcsin u)= 1/sqrt(1-u^2)*d/dx(u)# let me continue from the given #tan(arcsin x)# #d/dx(tan(arcsin x))=sec^2 (arcsin x)*d/dx(arcsin x)# #= sec^2 (arcsin x)*(1/sqrt(1-x^2))*d/dx(x)# the final answer is #= sec^2 (arcsin x)/sqrt(1-x^2)#

Elijah Abram · Answer

undefined To find the derivative of the function tan(arcsin x), you can use the chain rule. Let's denote y = arcsin x. Then, according to the chain rule, the derivative of tan(arcsin x) with respect to x is:

(dy/dx) = (dy/du) * (du/dx)

where u = arcsin x. Now, find the derivatives:

(dy/du) = d(tan u)/du = sec²(u)

(du/dx) = d(arcsin x)/dx = 1 / √(1 - x²)

Now, multiply these derivatives:

(dy/dx) = (sec²(u)) * (1 / √(1 - x²))

Substitute u = arcsin x:

(dy/dx) = sec²(arcsin x) / √(1 - x²)

Recall that secant squared of an angle θ is equal to 1 / cos²(θ). So, we can rewrite sec²(arcsin x) as 1 / cos²(arcsin x). And since cos(arcsin x) equals √(1 - x²), we have:

(dy/dx) = 1 / (cos²(arcsin x) * √(1 - x²))

Finally, using the trigonometric identity cos²(θ) = 1 - sin²(θ), we get:

(dy/dx) = 1 / (√(1 - sin²(arcsin x)) * √(1 - x²))

Simplify using the fact that sin(arcsin x) = x:

(dy/dx) = 1 / (√(1 - x²) * √(1 - x²))

(dy/dx) = 1 / (1 - x²)