Question #a62bb

Question

Question #a62bb

2 Answers

Impact of this question

1051 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-09T00:07:21

$\frac{d y}{d x} = 15 x^{2} {(1 + x^{3})}^{4}$ $dy/dx=15x^2(1+x^3)^4$

Explanation:

.

$y = {(1 + x^{3})}^{5}$ $y=(1+x^3)^5$

$u = 1 + x^{3}$ $u=1+x^3$

$\frac{d u}{d x} = 3 x^{2}$ $(du)/dx=3x^2$

$y = u^{5}$ $y=u^5$

$\frac{d y}{d u} = 5 u^{4}$ $dy/(du)=5u^4$

Using Chain Rule:

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $dy/dx=dy/(du)*(du)/dx$

$\frac{d y}{d x} = 5 u^{4} (3 x^{2})$ $dy/dx=5u^4(3x^2)$

$\frac{d y}{d x} = 5 {(1 + x^{3})}^{4} (3 x^{2})$ $dy/dx=5(1+x^3)^4(3x^2)$

$\frac{d y}{d x} = 15 x^{2} {(1 + x^{3})}^{4}$ $dy/dx=15x^2(1+x^3)^4$

Answer 2 · 2018-01-09T00:21:18

$f'(x)=15x^14+60x^11+90x^8+60x^5+15x^2$

Explanation:

You need to know two rules: the power rule and the chain rule.
The power rule states that if $f(x)=x^n$ , then $f'(x)=nx^(n-1)$
The chain rule states that if $f(x)=g(x)$ , then the derivative of $f(x)$ is the derivative of $g(x)$ times the derivative of $x$ .

Therefore, if $f(x)=(1+x^3)^5$ , then
$f'(x)=5(1+x^3)^4xx(3x^2)$ or
$f'(x)=15x^2(1+x^3)^4$
You could of course expand this to get:
$f'(x)=15x^14+60x^11+90x^8+60x^5+15x^2$

We're of course, saying that $f(x)=y$

Science

Math

Humanities

... and beyond

Question #a62bb

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question