## MTALHAHASSAN2 one year ago 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

1. MTALHAHASSAN2

@Shalante

2. anonymous

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)

3. MTALHAHASSAN2

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

4. MTALHAHASSAN2

is it right???

5. anonymous

$f(x) = (3x^2+4)(3+x^3)^5$$f(x) = 3x^2+4 ~,~ g(x) = (3+x^2)^5$

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

Remember to apply the chainrule.

10. anonymous

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

11. anonymous

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

12. anonymous

Ahh ok

13. MTALHAHASSAN2

=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

14. anonymous

Yes.