Question

Use the product rule and the power of a function rule to differenite the following functions. Do not simplify.
b) (3x^2+4)(3+x^3)^5

Answer 1

@Shalante

Answer 2

This can be solved using the product rule
Its just 2 function multiplying together: f(x) and g(x)

Lets call
\[f(x)=3x^2+4\]
\[g(x)=(3+x^3)^5\]

Try to find the derivative of both by yourself
f'(x)=?
g'(x)=?

then use this product rule formula
f'(x)g(x)+f(x)g'(x)

Answer 3

y=(3x^2+4)(3+x^3)^5
=2(3) (3+x^3)^5 + (3x^2+4) 15(3+x^2)

Answer 4

is it right???

Answer 5

\[f(x) = (3x^2+4)(3+x^3)^5\]\[f(x) = 3x^2+4 ~,~ g(x) = (3+x^2)^5\]

Answer 6

Derivative of 3x^2 is not 6
Derivative (3+x^3)^5 is not 15(3+x^2)

Answer 7

\[\frac{d}{dx} (f(x)) = \frac{d}{dx}(3x^2+4) \cdot (3+x^3)^5 +\frac{d}{dx}((3+x^3)^5) \cdot (3x^2+4)\]

Answer 8

You forgot the x in the first one
You forgot the x^2 in the second one.

Answer 9

Remember to apply the chainrule.

Answer 10

\[\frac{d}{dx}(f(g(x)) = f'(g(x)) \cdot g'(x)\]

Answer 11

He did apply the chain rule.
He forgot the x on both cases but not the numbers.

Answer 12

Ahh ok

Answer 13

=2(3x)(3+x^3)^5+(3x^2+4)5(3x^2)(3+x^3)^4
=6x(3+x^3)^5+15x^2(3x^2+4)(3+x^3)^4

Answer 14

Yes.