Question

Calc help, derivative 25arcsin(x/5) -xsqrt(25-x^(2))

Answer 1

If possible help step by step

Answer 2

Possible derivation:
d/dx(25 sin^(-1)(x/5)-x sqrt(25-x^2))
| Differentiate the sum term by term and factor out constants:
= | 25 (d/dx(sin^(-1)(x/5)))-d/dx(x sqrt(25-x^2))
| Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x and v = sqrt(25-x^2):
= | 25 (d/dx(sin^(-1)(x/5)))-(sqrt(25-x^2) (d/dx(x))+x (d/dx(sqrt(25-x^2))))
| Use the chain rule, d/dx(sin^(-1)(x/5)) = ( dsin^(-1)(u))/( du) ( du)/( dx), where u = x/5 and ( dsin^(-1)(u))/( du) = 1/sqrt(1-u^2):
= | -x (d/dx(sqrt(25-x^2)))+25 (d/dx(x/5))/sqrt(1-x^2/25)-sqrt(25-x^2) (d/dx(x))
| The derivative of x is 1:
= | -x (d/dx(sqrt(25-x^2)))+(25 (d/dx(x/5)))/sqrt(1-x^2/25)-sqrt(25-x^2)
| Factor out constants:
= | (25 (1/5 (d/dx(x))))/sqrt(1-x^2/25)-x (d/dx(sqrt(25-x^2)))-sqrt(25-x^2)
| Use the chain rule, d/dx(sqrt(25-x^2)) = ( dsqrt(u))/( du) ( du)/( dx), where u = 25-x^2 and ( dsqrt(u))/( du) = 1/(2 sqrt(u)):
= | -x (d/dx(25-x^2))/(2 sqrt(25-x^2))+(5 (d/dx(x)))/sqrt(1-x^2/25)-sqrt(25-x^2)
| The derivative of x is 1:
= | -(x (d/dx(25-x^2)))/(2 sqrt(25-x^2))-sqrt(25-x^2)+5/sqrt(1-x^2/25)
| Differentiate the sum term by term and factor out constants:
= | -(x (d/dx(25)-d/dx(x^2)))/(2 sqrt(25-x^2))-sqrt(25-x^2)+5/sqrt(1-x^2/25)
| The derivative of 25 is zero:
= | -(x (0-d/dx(x^2)))/(2 sqrt(25-x^2))-sqrt(25-x^2)+5/sqrt(1-x^2/25)
| The derivative of x^2 is 2 x:
= | (x (2 x))/(2 sqrt(25-x^2))-sqrt(25-x^2)+5/sqrt(1-x^2/25)

Answer 3

thanks this makes more sense