tentukan turunan dari fungsi tersebut, dari gambar yang telah di unggah
Tentukan turunan dari fungsi-fungsi berikut dari gambar yang diunggah.
I can see the mathematical functions listed in the image, and I will calculate their derivatives. Here are the functions and their respective derivatives:
y = x 2 + 5 x y = x^2 + 5x y = x 2 + 5 x
Derivative:
y ′ = d d x ( x 2 ) + d d x ( 5 x ) = 2 x + 5 y' = \frac{d}{dx}(x^2) + \frac{d}{dx}(5x) = 2x + 5 y ′ = d x d ( x 2 ) + d x d ( 5 x ) = 2 x + 5
y = 3 x 4 + x 3 − 2 x y = 3x^4 + x^3 - 2x y = 3 x 4 + x 3 − 2 x
Derivative:
y ′ = d d x ( 3 x 4 ) + d d x ( x 3 ) − d d x ( 2 x ) = 12 x 3 + 3 x 2 − 2 y' = \frac{d}{dx}(3x^4) + \frac{d}{dx}(x^3) - \frac{d}{dx}(2x) = 12x^3 + 3x^2 - 2 y ′ = d x d ( 3 x 4 ) + d x d ( x 3 ) − d x d ( 2 x ) = 12 x 3 + 3 x 2 − 2
y = 3 x 4 x 5 y = \frac{3x}{4x^5} y = 4 x 5 3 x
Simplify first:
y = 3 4 x 4 y = \frac{3}{4x^4} y = 4 x 4 3
Derivative:
y ′ = d d x ( 3 4 x 4 ) = − 3 x − 5 y' = \frac{d}{dx}\left( \frac{3}{4x^4} \right) = -3x^{-5} y ′ = d x d ( 4 x 4 3 ) = − 3 x − 5
y = x 2 + 2 x + 7 y = \frac{x^2 + 2}{x + 7} y = x + 7 x 2 + 2
Use the quotient rule:
y ′ = ( x + 7 ) ( 2 x ) − ( x 2 + 2 ) ( 1 ) ( x + 7 ) 2 = 2 x ( x + 7 ) − ( x 2 + 2 ) ( x + 7 ) 2 y' = \frac{(x + 7)(2x) - (x^2 + 2)(1)}{(x + 7)^2} = \frac{2x(x + 7) - (x^2 + 2)}{(x + 7)^2} y ′ = ( x + 7 ) 2 ( x + 7 ) ( 2 x ) − ( x 2 + 2 ) ( 1 ) = ( x + 7 ) 2 2 x ( x + 7 ) − ( x 2 + 2 ) y ′ = 2 x 2 + 14 x − x 2 − 2 ( x + 7 ) 2 = x 2 + 14 x − 2 ( x + 7 ) 2 y' = \frac{2x^2 + 14x - x^2 - 2}{(x + 7)^2} = \frac{x^2 + 14x - 2}{(x + 7)^2} y ′ = ( x + 7 ) 2 2 x 2 + 14 x − x 2 − 2 = ( x + 7 ) 2 x 2 + 14 x − 2
y = ( x 2 + 2 ) ( x 3 + 1 ) y = (x^2 + 2)(x^3 + 1) y = ( x 2 + 2 ) ( x 3 + 1 )
Use the product rule:
y ′ = ( x 2 + 2 ) ⋅ d d x ( x 3 + 1 ) + ( x 3 + 1 ) ⋅ d d x ( x 2 + 2 ) y' = (x^2 + 2) \cdot \frac{d}{dx}(x^3 + 1) + (x^3 + 1) \cdot \frac{d}{dx}(x^2 + 2) y ′ = ( x 2 + 2 ) ⋅ d x d ( x 3 + 1 ) + ( x 3 + 1 ) ⋅ d x d ( x 2 + 2 ) y ′ = ( x 2 + 2 ) ( 3 x 2 ) + ( x 3 + 1 ) ( 2 x ) = 3 x 4 + 6 x 2 + 2 x 4 + 2 x y' = (x^2 + 2)(3x^2) + (x^3 + 1)(2x) = 3x^4 + 6x^2 + 2x^4 + 2x y ′ = ( x 2 + 2 ) ( 3 x 2 ) + ( x 3 + 1 ) ( 2 x ) = 3 x 4 + 6 x 2 + 2 x 4 + 2 x y ′ = 5 x 4 + 6 x 2 + 2 x y' = 5x^4 + 6x^2 + 2x y ′ = 5 x 4 + 6 x 2 + 2 x
y = x ( x 2 + 1 ) y = x(x^2 + 1) y = x ( x 2 + 1 )
Use the product rule:
y ′ = x ⋅ d d x ( x 2 + 1 ) + ( x 2 + 1 ) ⋅ d d x ( x ) = x ( 2 x ) + ( x 2 + 1 ) ( 1 ) y' = x \cdot \frac{d}{dx}(x^2 + 1) + (x^2 + 1) \cdot \frac{d}{dx}(x) = x(2x) + (x^2 + 1)(1) y ′ = x ⋅ d x d ( x 2 + 1 ) + ( x 2 + 1 ) ⋅ d x d ( x ) = x ( 2 x ) + ( x 2 + 1 ) ( 1 ) y ′ = 2 x 2 + x 2 + 1 = 3 x 2 + 1 y' = 2x^2 + x^2 + 1 = 3x^2 + 1 y ′ = 2 x 2 + x 2 + 1 = 3 x 2 + 1
y = 3 x ( x 3 − 1 ) y = 3x(x^3 - 1) y = 3 x ( x 3 − 1 )
Use the product rule:
y ′ = 3 x ⋅ d d x ( x 3 − 1 ) + ( x 3 − 1 ) ⋅ d d x ( 3 x ) y' = 3x \cdot \frac{d}{dx}(x^3 - 1) + (x^3 - 1) \cdot \frac{d}{dx}(3x) y ′ = 3 x ⋅ d x d ( x 3 − 1 ) + ( x 3 − 1 ) ⋅ d x d ( 3 x ) y ′ = 3 x ( 3 x 2 ) + ( x 3 − 1 ) ( 3 ) = 9 x 3 + 3 x 3 − 3 y' = 3x(3x^2) + (x^3 - 1)(3) = 9x^3 + 3x^3 - 3 y ′ = 3 x ( 3 x 2 ) + ( x 3 − 1 ) ( 3 ) = 9 x 3 + 3 x 3 − 3 y ′ = 12 x 3 − 3 y' = 12x^3 - 3 y ′ = 12 x 3 − 3
Let me know if you would like further explanation on any specific derivative.
Follow-up Questions:
What is the quotient rule for derivatives, and how is it applied?
How can we simplify complex rational functions before differentiating?
Can you explain the product rule in detail and give more examples?
How would the chain rule be used if the functions were composed?
What are higher-order derivatives and when are they useful?
Tip: Always simplify a function as much as possible before differentiating to make the process easier and reduce mistakes.