Question

Find the derivative of the function.

1) f(x) = sin^-1(4x) sin is inverse sin
2) y = tan^-1(√(x-2)) tan is inverse tan
3) y = (tan^−1(5x))2 tan is inverse tan
4) h(x) = e^(x^8+ln(x))
5) y = ln(e^−x + xe^−x)
6) y = (cos(9x))^x

Answer 1

Explanation:

f(x) = sin^-1(4x)

Using the chain rule, we have:

f'(x) = (1/√(1 - (4x)^2)) * d/dx(4x)

= (1/√(1 - (4x)^2)) * 4

= 4/√(1 - (4x)^2)

y = tan^-1(√(x-2))

Using the chain rule, we have:

y' = (1/ (1 + (√(x-2))^2)) * (1/2)(x-2)^(-1/2)

= 1 / (2√(x-2)(1 + x - 2))

= 1 / (2√(x-2)(x - 1))

y = (tan^−1(5x))^2

Using the chain rule and power rule, we have:

y' = 2(tan^−1(5x)) * d/dx(tan^−1(5x))

= 2(tan^−1(5x)) * (1/(1 + (5x)^2)) * 5

= 10(tan^−1(5x))/(1 + (5x)^2)

h(x) = e^(x^8+ln(x))

Using the chain rule and product rule, we have:

h'(x) = e^(x^8+ln(x)) * d/dx(x^8+ln(x))

= e^(x^8+ln(x)) * (8x^7 + 1/x)

= x^(15)e^(x^8) + e^(x^8)/x

y = ln(e^−x + xe^−x)

Using the chain rule and sum rule, we have:

y' = (1/(e^−x + xe^−x)) * d/dx(e^−x + xe^−x)

= (1/(e^−x + xe^−x)) * (-e^−x + e^−x - xe^−x)

= -1/(e^−x + xe^−x)

y = (cos(9x))^x

Using the chain rule and power rule, we have:

y' = (cos(9x))^x * d/dx(x * ln(cos(9x)))

= (cos(9x))^x * (ln(cos(9x)) + x * (-sin(9x)) * (1/cos(9x)) * 9)

= (cos(9x))^x * (ln(cos(9x)) - 9x * tan(9x))