Tentukan turunanya (1) $x^{3}+y^{3}-3 x y=0$ (2) $(3 y)^{ {1}{3}}+(4 y)^{ {1}{4}}=x$ (3) $ {x+y}+ {x y}= {6}$ (9) $ {y}{x}+ {x}{y}=4$ (- $ {1}{x}+ {1}{y}=1$ (6) $y=x^{2} x$ (c) $s= ^{3} {1}{t}+ ^{3} {1}{t}$ (8) $v= ( { u})$ (0) $f(x)= {x}{ x+ x}$ (10) $ (x+y)=y x$See answer

Tentukan turunanya 1 x3 y3 3 x y 0 2 3 y 13 4 y 14 x 3 x y x y 6 9 yx xy 4 1x 1y 1 6 y x2 x c s 3 1t 3 1t 8 v u 0 f x x x x 10 x y y x

Question

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Tentukan turunanya (1) $x^{3}+y^{3}-3 x y=0$ (2) $(3 y)^{\frac{1}{3}}+(4 y)^{\frac{1}{4}}=x$ (3) $\sqrt{x+y}+\sqrt{x y}=\sqrt{6}$ (9) $\frac{y}{x}+\frac{x}{y}=4$ (- $\frac{1}{x}+\frac{1}{y}=1$ (6) $y=x^{2} \sin x$ (c) $s=\sin ^{3} \frac{1}{t}+\cos ^{3} \frac{1}{t}$ (8) $v=\sin (\tan \sqrt{\sin u})$ (0) $f(x)=\frac{x}{\sin x+\cos x}$ (10) $\cos (x+y)=y \sin x$

Basic Answer

The image contains 10 questions in Indonesian, asking for the derivative of various functions. I will solve them one by one. Note that implicit differentiation will be used where necessary.

Step 1: Question 1 – Implicit Differentiation

The equation is x³ + y³ – 3xy = 0. Differentiate both sides with respect to x:

3x² + 3y²(dy/dx) – 3y – 3x(dy/dx) = 0

Solve for dy/dx:

dy/dx (3y² – 3x) = 3y – 3x²

dy/dx = (3y – 3x²) / (3y² – 3x) = (y – x²) / (y² – x)

Step 2: Question 2 – Implicit Differentiation

The equation is (3y)^(1/3) + (4y)^(1/4) = x. Differentiate both sides with respect to x:

(1/3)(3y)^(-2/3) * 3(dy/dx) + (1/4)(4y)^(-3/4) * 4(dy/dx) = 1

(3y)^(-2/3)(dy/dx) + (4y)^(-3/4)(dy/dx) = 1

dy/dx [(3y)^(-2/3) + (4y)^(-3/4)] = 1

dy/dx = 1 / [(3y)^(-2/3) + (4y)^(-3/4)]

Step 3: Question 3 – Implicit Differentiation

The equation is √x + y + √(xy) = √6. Differentiate both sides with respect to x:

(1/2)x^(-1/2) + dy/dx + (1/2)(xy)^(-1/2) * (y + x(dy/dx)) = 0

Solve for dy/dx (this will be a complex expression).

Step 4: Question 4 – Implicit Differentiation

The equation is y/x + x/y = 4. Differentiate both sides with respect to x:

(x(dy/dx) – y)/x² + (y – x(dy/dx))/y² = 0

Solve for dy/dx.

Step 5: Question 5 – Implicit Differentiation

The equation is 1/x + 1/y = 1. Differentiate both sides with respect to x:

-1/x² – (1/y²)(dy/dx) = 0

dy/dx = -y²/x²

Step 6: Question 6 – Product Rule

The equation is y = x²sin(x). Using the product rule:

dy/dx = 2xsin(x) + x²cos(x)

Step 7: Question 7 – Chain Rule

The equation is s = sin³(1/t) + cos³(1/t). Using the chain rule:

ds/dt = 3sin²(1/t)cos(1/t)(-1/t²) + 3cos²(1/t)(-sin(1/t))(-1/t²)

Step 8: Question 8 – Chain Rule

The equation is v = sin(tan(√sin(u))). Using the chain rule (multiple times):

dv/du = cos(tan(√sin(u))) * sec²(√sin(u)) * (1/2)(sin(u))^(-1/2) * cos(u)

Step 9: Question 9 – Quotient Rule

The equation is f(x) = x / (sin(x) + cos(x)). Using the quotient rule:

f'(x) = [(sin(x) + cos(x)) – x(cos(x) – sin(x))] / (sin(x) + cos(x))²

Step 10: Question 10 – Implicit Differentiation

The equation is cos(x + y) = ysin(x). Differentiate both sides with respect to x:

-sin(x + y)(1 + dy/dx) = (dy/dx)sin(x) + ycos(x)