Please help me.

Question

1. Use the definition of the derivative to find f'(x), if f(x)=x^-2

2.Find the derivative of y=3t^5 - 5√t + 7/t

Answer 1

Use the form:
f'(x^n) = nx^(n-1)

Answer 2

1.

f ( x ) = x ^ - 2 = 1 / x ^ 2

f'(x) = lim Δ f ( x ) / Δx
Δx ->0

Δ f ( x ) = 1 / ( x + Δx ) ^ 2 - 1 / x ^ 2 =

[ x ^ 2 - ( x + Δx ) ^ 2 ] / [ ( x + Δx ) ^ 2 * x ^ 2 ] =

[ x ^ 2 - ( x ^ 2 + 2 x * Δx + Δx ^ 2 ) ] / [ ( x + Δx ) ^ 2 * x ^ 2 ] =

( x ^ 2 - x ^ 2 - 2 x * Δx - Δx ^ 2 ) / [ ( x + Δx ) ^ 2 * x ^ 2 ] =

( - 2 x * Δx - Δx ^ 2 ) / [ ( x + Δx ) ^ 2 * x ^ 2 ] =

- Δx * ( 2 x + Δx ) / [ ( x + Δx ) ^ 2 * x ^ 2 ]

Δ f ( x ) / Δx = { - Δx * ( 2 x + Δx ) / [ ( x + Δx ) ^ 2 * x ^ 2 ] } / Δx =

- Δx * ( 2 x + Δx ) / [ Δx * ( x + Δx ) ^ 2 * x ^ 2 ] =

- ( 2 x + Δx ) / [ ( x + Δx ) ^ 2 * x ^ 2 ]

When Δx->0 then:

- ( 2 x + Δx ) = - ( 2 x + 0 ) = - 2 x

( x + Δx ) ^ 2 = ( x + 0 ) ^ 2 = x ^ 2

( x + Δx ) ^ 2 * x ^ 2 = ( x + 0 ) ^ 2 * x ^ 2 =

x ^ 2 * x ^ 2 = x ^ 4

so

f'(x) = lim Δ f ( x ) / Δx =
Δx ->0

lim - ( 2 x + Δx ) / [ ( x + Δx ) ^ 2 * x ^ 2 ] =
Δx ->0

- 2 x / x ^ 4 =

- 2 / x ^ 3

f' ( x ^ - 2 ) = - 2 / x ^ 3

2.

y = 3 t ^ 5 - 5 √t + 7 / t =

3 t ^ 5 - 5 * t ^ ( 1 / 2 ) + 7 * t ^ - 1

Use the power rule : ( t ^ n ) ´ = n * t ^ ( n - 1 )

y ´ = ( 3 t ^ 5 - 5 * t ^ ( 1 / 2 ) + 7 * t ^ - 1 ) ´ =

3 * 5 t ^ ( 5 - 1 ) - 5 * ( 1 / 2 ) * t ( 1 / 2 - 1 ) + 7 * ( - 1 ) * t ^ ( - 1 - 1 ) =

15 t ^ 4 - ( 5 / 2 ) * t ( - 1 / 2 ) - 7 * t ^ - 2 =

15 t ^ 4 - 7 / t ^ 2 - 5 / 2√t