How do you find the derivative of $(tan^3(x)+10)^2$ ?

Question

1917 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-04T09:45:08

I get $6tan^2(x)sec^2(x)(tan^3(x)+10)$ .

We use the chain rule, which states that

$(df)/dx=(df)/(du)*(du)/dx$

Let $u=tan^3(x)+10$ , so we have $f=u^2$ , then $(df)/(du)=2u$

We can also differentiate $tan^3(x)$ using the chain rule.

We let $z=tan(x),f=z^3$ , then $(df)/dz=3z^2$ , and $(dz)/dx=sec^2(x)$ .

So, the derivative of $tan^3(x)$ is $3z^2sec^2(x)=3tan^2(x)sec^2(x)$ .

That is also the $(du)/dx$ .

Putting everything back together, we get

$(df)/dx=2u*3tan^2(x)sec^2(x)$

Replacing back $u=tan^3(x)+10$ , we get

$(df)/dx=2(tan^3(x)+10)*3tan^2(x)sec^2(x)$

$=6tan^2(x)sec^2(x)(tan^3(x)+10)$

How do you find the derivative of (tan^3(x)+10)^2 (tan^3(x)+10)^2?