## Calculating Derivatives: Problems and Solutions

### Calculating Derivatives: Problems and Solutions

Are you working to calculate derivatives in Calculus? Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself.

Jump down this page to: [Power rule, $x^n$] [Exponential, $e^x$] [Trig derivatives] [Product rule] [Quotient rule] [Chain rule]

Two specific cases you’ll quickly remember:

$$\dfrac{d}{dx}\text{(constant)} = 0$$

$$\dfrac{d}{dx}(x) = 1$$

Recall $\sqrt[n]{x} = x^{1/n}.$

Recall $\dfrac{1}{x^n} = x^{-n}.$

This one’s easy to remember!

\begin{align*}

\frac{d}{dx}\left(\sin x\right) &= \cos x &&& \frac{d}{dx}\left(\csc x\right) &= -\csc x \cot x \\ \\

\dfrac{d}{dx}\left(\cos x\right) &= -\sin x &&& \frac{d}{dx}\left(\sec x\right) &= \sec x \tan x \\ \\

\dfrac{d}{dx}\left(\tan x\right) &= \sec^2 x &&& \frac{d}{dx}\left(\cot x\right) &= -\csc^2 x

\end{align*}} \] Notice that a negative sign appears in the derivatives of the co-functions: cosine, cosecant, and cotangent.

\begin{align*}

\dfrac{d}{dx}(fg)&= \left(\dfrac{d}{dx}f \right)g + f\left(\dfrac{d}{dx}g \right)\\[8px] &= [{\small \text{ (deriv of the 1st) } \times \text{ (the 2nd) }}]\, + \,[{\small \text{ (the 1st) } \times \text{ (deriv of the 2nd)}}] \end{align*}}\]

\begin{align*}

\dfrac{d}{dx}\left(\dfrac{f}{g} \right) &= \dfrac{\left(\dfrac{d}{dx}f \right)g – f\left(\dfrac{d}{dx}g \right)}{g^2} \\[8px] &=\dfrac{{[{\small \text{(derivative of the numerator) } \times \text{ (the denominator)}]}\\ \quad – \, [{\small \text{ (the numerator) } \times \text{ (derivative of the denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}}

\end{align*}}\] Many students remember the quotient rule by thinking of the numerator as “hi,” the demoninator as “lo,” the derivative as “d,” and then singing

“lo d-hi minus hi d-lo over lo-lo”

Have a question, suggestion, or item you’d like us to include? Please let us know in the Comments section below!

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In Power Rule Differentiation Problem #3, please explain why d/dx(33)^0 is striked out?

Thanks for asking! That’s just a convenient way to show that the derivative of that term, the constant 33, is 0: d/dx(33) = 0. Any time you have a value that equals zero in an equation, drawing a slanted arrow that points to 0 can help the reader (your teacher, or whomever) see the reason that you’re not carrying the term on to the next line.

Hope that helps, and please let us know any other questions that you have! : )

I love this idea , and the solution is very good i mean its easy to understand

Thanks for letting us know, Ibrahim. That’s our aim: to make things easy for you to understand and then be able to do yourself!

It was really helpful…..thanks for the given solutions which made understanding the topic easier

Thanks, Pratistha! We’re glad to know that our solutions made understanding how to calculate derivatives easier for you. : )

It’s very much helpful

Thanks for writing to tell us. We’ve very happy to know this was useful to you! : )