How do I find the derivative of x^2 with respect to x in the following question?

Question

I apologize in advance if I use an image to present my problem. I'm having a tough time putting it on here. I know how to complete part a of the problem, but I am stumped on parts b and c. Any suggestions?

Answer 1

For $b$, you must use the power rule for derivatives which states

$$ \frac{d}{dx} x^n = nx^{n-1}$$

Hence,

$$ \frac{d}{dx} x^2 = 2x^{2-1} = 2x^{1} = 2x$$

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For $c$, this is easier to understand with a substitution as follows: let $u = x^2$. Then,

$$ \frac{dy}{dx^2} = \frac{d}{dx^2} (8x^4 - 6x^2 +x - 1) = \frac{d}{du} (8u^2 - 6u + \sqrt{u} - 1)$$

Now you can solve normally, and substitute back to $x$ when finished. By the sum/difference rule for derivatives,

$$ \frac{d}{du} (8u^2 - 6u + \sqrt{u} - 1) = \frac{d}{du} 8u^2 - \frac{d}{du} 6u + \frac{d}{du} \sqrt{u} - \frac{d}{du} 1$$

By the constant multiple rule for derviatives,

$$ 8\frac{d}{du} u^2 - 6\frac{d}{du} u + \frac{d}{du} \sqrt{u} - \frac{d}{du} 1 $$

By the constant rule for derivatives,

$$ 8\frac{d}{du} u^2 - 6\frac{d}{du} u + \frac{d}{du} \sqrt{u} - 0 $$

Rewrite square root as an exponent so you can easily use the power rule...

$$ 8\frac{d}{du} u^2 - 6\frac{d}{du} u + \frac{d}{du} u^{\frac{1}{2}} $$

Now use the power rule for derivatives,

$$ 8 \cdot 2u - 6 \cdot 1 + \frac{1}{2} u^{\frac{-1}{2}} $$

Finish with basic algebra,

$$ 16u - 6 + \frac{1}{2 \sqrt{u}} $$

Substitute back in $x$

$$ 16x^2 - 6 + \frac{1}{2 \sqrt{x^2}} = 16x^2 - 6 + \frac{1}{2x}$$

Answer 2

You actually already took the derivative in part a and multiplied it by $-6$. Because you computed:

$$\frac{d}{dx}(y)$$

or

$$\frac{d}{dx}\left(8x^4-6x^2+x-1\right)$$

or

$$8\left(\frac{d}{dx}(x^4)\right)-6\left(\frac{d}{dx}(x^2)\right)+\frac{d}{dx}(x)-\frac{d}{dx}(1)$$

When you computed $\frac{dy}{dx}$ where $y=8x^4-6x^2+x-1$, you were taking the derivative of $y$ with respect to $x$ or $\frac{d}{dx}(y)$. Use the same technique to take the derivative of $x^2$ with respect to $x$ or $\frac{d}{dx}(x^2)$.