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High School Math : Finding Derivatives

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Example Questions

Example Question #2 : Understanding Derivatives Of Sums, Quotients, And Products

What is the first derivative of $5x^4+\sin x$ $\displaystyle 5x^4+\sin x$?

Possible Answers:

$20x^3+\frac{1}{\sin x}$ $\displaystyle 20x^3+\frac{1}{\sin x}$

$20x^{3}-\cos x$ $\displaystyle 20x^{3}-\cos x$

$5x^4\cos x +20x^3\sin x$ $\displaystyle 5x^4\cos x +20x^3\sin x$

$20x^3+4\cos x$ $\displaystyle 20x^3+4\cos x$

$20x^{3}+\cos x$ $\displaystyle 20x^{3}+\cos x$

Correct answer:

$20x^{3}+\cos x$ $\displaystyle 20x^{3}+\cos x$

Explanation:

Since we're adding terms, we take the derivative of each part separately. For 5x^4 $\displaystyle 5x^4$, we can use the power rule, which states that we multiply the variable by the current exponent and then lower the exponent by one. For sine, we use our trigonometric derivative rules.

Remember, $\frac{\mathrm{d} }{\mathrm{d} x} \sin x=\cos x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin x=\cos x$.

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(4*5x^{4-1})+\cos x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(4*5x^{4-1})+\cos x$

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(20x^{3})+\cos x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(20x^{3})+\cos x$

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Example Question #61 : Finding Derivatives

What is the second derivative of $5x^4+\sin x$ $\displaystyle 5x^4+\sin x$?

Possible Answers:

$60x^{2}-\sin x$ $\displaystyle 60x^{2}-\sin x$

$40x^{2}-\sin x$ $\displaystyle 40x^{2}-\sin x$

$120x+\sin x$ $\displaystyle 120x+\sin x$

$40x^2-\cos x$ $\displaystyle 40x^2-\cos x$

$x^5-\csc x$ $\displaystyle x^5-\csc x$

Correct answer:

$60x^{2}-\sin x$ $\displaystyle 60x^{2}-\sin x$

Explanation:

To find the second derivative, we need to start by finding the first one.

Remember, $\frac{\mathrm{d} }{\mathrm{d} x} \sin x=\cos x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin x=\cos x$.

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(4*5x^{4-1})+\cos x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(4*5x^{4-1})+\cos x$

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(20x^{3})+\cos x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(20x^{3})+\cos x$

Now we repeat the process, but using $20x^3+\cos x$ $\displaystyle 20x^3+\cos x$ as our equation.

$\frac{\mathrm{d} }{\mathrm{d} x} \cos x=-\sin x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos x=-\sin x$

$\frac{\mathrm{d} }{\mathrm{d} x}20x^3+\cos x=(3*20x^{3-1})-\sin x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}20x^3+\cos x=(3*20x^{3-1})-\sin x$

$\frac{\mathrm{d} }{\mathrm{d} x}20x^3+\cos x=(60x^{2})-\sin x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}20x^3+\cos x=(60x^{2})-\sin x$

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Example Question #62 : Finding Derivatives

Which of the following best represents $\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}$?

Possible Answers:

$\displaystyle f'(x)g(x)+g'(x)f(x)$

$\frac{f'(x)}{g'(x)}$ $\displaystyle \frac{f'(x)}{g'(x)}$

$\frac{f(x)g'(x)-g(x)f'(x)}{g(x)^2}$ $\displaystyle \frac{f(x)g'(x)-g(x)f'(x)}{g(x)^2}$

$\frac{g(x)g'(x)-f(x)f'(x)}{g(x)^2}$ $\displaystyle \frac{g(x)g'(x)-f(x)f'(x)}{g(x)^2}$

$\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ $\displaystyle \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

Correct answer:

$\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ $\displaystyle \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

Explanation:

The question is just asking for the Quotient Rule formula.

Recall the Quotient Rule is the bottom function times the derivative of the top minus the top function times the derivative of the bottom all divided by the bottom function squared.

Given,

$\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}$

the bottom function is g(x) $\displaystyle g(x)$ and the top function is f(x) $\displaystyle f(x)$. This makes the bottom derivative g'(x) $\displaystyle g'(x)$ and the top derivative f'(x) $\displaystyle f'(x)$.

Substituting these into the Quotient Rule formula resulting in the following.

$\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}=\frac{g(x)\cdot f'(x)-f(x)\cdot g'(x)}{(g(x))^2}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}=\frac{g(x)\cdot f'(x)-f(x)\cdot g'(x)}{(g(x))^2}$

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High School Math : Finding Derivatives

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #2 : Understanding Derivatives Of Sums, Quotients, And Products

Example Question #61 : Finding Derivatives

Example Question #62 : Finding Derivatives

All High School Math Resources

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