first derivative of second * second derivative of first
(x²+2)³ 2(x^3+3)(3x^2)
+
(x³+3)² 3(x^2+2)^2(2x)
which is
6x^2(x^2+2)^3(x^3+3)+6x(x^2+2)^2(x^3+3)^2
which is D
A.6(x²+2)²(x³+3)
B.2x(x³+3)² +3x² (x² +2)³
C.3(x²+2)²(x³+3)²+2(x²+2)²(x³+3)
D.6x(x²+2)²(x³+3)²+6x²(x²+2)³(x³+3)
D or C I'm not sure ??
(x²+2)³ 2(x^3+3)(3x^2)
+
(x³+3)² 3(x^2+2)^2(2x)
which is
6x^2(x^2+2)^3(x^3+3)+6x(x^2+2)^2(x^3+3)^2
which is D
y'=d/dx 6u^3 v^2=6*3(u^2)vdu/dx+12u^3vdv/dx
where u= (x^2+2) du= 2x dx
v= x^3+3 dv=3x^2 dx
can you take it from there?
Let's break down the function first:
y = (x²+2)³(x³+3)²
Applying the product rule, the derivative of y would be:
dy/dx = [(x³+3)²]*(d/dx)(x²+2)³ + [(x²+2)³]*(d/dx)(x³+3)²
Now, let's calculate the derivatives of each term separately:
Term 1: (d/dx)(x²+2)³
To find the derivative of (x²+2)³, we need to apply the chain rule. The chain rule states that the derivative of f(g(x)) = f'(g(x)) * g'(x). In this case, f(u) = u³ and g(x) = x²+2.
So, f'(u) = 3u² (power rule for differentiation) and g'(x) = 2x (derivative of x²+2).
Applying the chain rule, the derivative of (x²+2)³ would be:
= 3(x²+2)² * 2x
= 6x(x²+2)²
Term 2: (d/dx)(x³+3)²
Similar to the previous term, to find the derivative of (x³+3)², we again apply the chain rule with f(u) = u² and g(x) = x³+3.
So, f'(u) = 2u (power rule for differentiation) and g'(x) = 3x² (derivative of x³+3).
Applying the chain rule, the derivative of (x³+3)² would be:
= 2(x³+3) * 3x²
= 6x²(x³+3)
Now, substituting these derivative results back into the product rule equation for dy/dx:
dy/dx = [(x³+3)²] * (6x(x²+2)²) + [(x²+2)³] * (6x²(x³+3))
Simplifying this expression further, we get:
dy/dx = 6(x²+2)²(x³+3)² + 6x²(x²+2)³(x³+3)
Therefore, option D is the correct answer.
So, the answer to the question "What is dy/dx if y=(x²+2)³(x³+3)²?" is D. 6x(x²+2)²(x³+3)²+6x²(x²+2)³(x³+3).