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

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

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Answer 1 · 2018-03-04T09:45:08

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

Explanation:

We use the chain rule, which states that

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

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

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

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

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

That is also the $\frac{d u}{d x}$ $(du)/dx$ .

Putting everything back together, we get

$\frac{d f}{d x} = 2 u \cdot 3 {tan}^{2} (x) {sec}^{2} (x)$ $(df)/dx=2u*3tan^2(x)sec^2(x)$

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

$\frac{d f}{d x} = 2 ({tan}^{3} (x) + 10) \cdot 3 {tan}^{2} (x) {sec}^{2} (x)$ $(df)/dx=2(tan^3(x)+10)*3tan^2(x)sec^2(x)$

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

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

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How do you find the derivative of (tan3(x)+10)2 (tan^3(x)+10)^2?

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Explanation:

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Impact of this question

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