Calculus Archives - Epsilonify

Derivative of sin^4(x)

The Mathematician — Fri, 09 Jun 2023 13:00:03 +0000

What is the derivative of sin^4(x)?

The derivative of $\sin^4(x)$ is $4\sin^3(x)\cos(x)$ .

Solution of the derivative of sin^4(x)

Solution: let $F(x) = g(f(x)) = \sin^4(x)$ where $g(u) = u^4$ and $f(x) = \sin(x)$ . To find the derivative of $\sin^4(x)$ , we need to apply the chain rule on $F(x)$ :

\begin{equation*} F'(x) = g'(f(x))f'(x) \end{equation*}

The derivative of $u^4$ is $4u^3$ and the derivative of $\sin(x)$ is $\cos(x)$ , which we have seen here earlier. Therefore, we get:

\begin{equation*} g'(f(x)) = g'(\sin(x)) = 4\sin^3(x) \text{ and } f'(x) = \cos(x). \end{equation*}

Substituting everything, we get:

\begin{align*}F'(x) &= g'(f(x))f'(x) \\&= 4\sin^3(x)\cos(x)\end{align*}

Therefore, the derivative of $\sin^4(x)$ is $4\sin^3(x)\cos(x)$ .

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Integral of 1/(x^2 – a^2)

The Mathematician — Mon, 05 Jun 2023 13:00:39 +0000

What is the integral of 1/(x^2 – a^2)?

The integral of

\frac{1}{x^2 - a^2}

\frac{1}{2a}\ln\lvert \frac{x - a}{x + a} \rvert + C

Solution of the integral 1/(x^2 – a^2)

Solution: We want to determine the next integral:

\begin{align*} \int \frac{dx}{x^2 - a^2}. \end{align*}

We will apply an integral technique such that the denominators will be linear functions:

\begin{align*} \frac{1}{x^2 - a^2} &= \frac{1}{(x-a)(x+a)} \\ &= \frac{A}{x - a} + \frac{B}{x + a} \\ &= \frac{Ax + Aa + Bx - Ba}{x^2 - a^2}. \end{align*}

We do have the following set of equalities:

\begin{align*} A + B &= 0 \\ Aa - Ba &= 1. \end{align*}

We see that

A = -B,

and therefore

B = -\frac{1}{2a}

and

A = \frac{1}{2a}

. Substituting everything above, we get:

\begin{align*} \int \frac{1}{x^2 - a^2} dx &= \int \frac{A}{x - a} dx + \int \frac{B}{x + a} dx \\ &= \frac{1}{2a} \int \frac{dx}{x - a} - \frac{1}{2a} \int \frac{dx}{x + a} \\ &= \frac{1}{2a} \ln \lvert x - a \rvert - \frac{1}{2a} \ln \lvert x + a \rvert + C \\ &= \frac{1}{2a} \ln \bigg\lvert \frac{x - a}{x + a} \bigg\rvert + C. \end{align*}

Conclusion

In conclusion, the integral of

\frac{1}{x^2 - a^2}

\frac{1}{2a}\ln \lvert \frac{x - a}{x + a} \rvert + C

, or in mathematical notation:

\begin{equation*} \int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln \bigg\lvert \frac{x - a}{x + a} \bigg\rvert + C. \end{equation*}The post Integral of 1/(x^2 – a^2) appeared first on Epsilonify.

Integral of 1/(x^2 + a^2)

The Mathematician — Tue, 30 May 2023 13:00:46 +0000

What is the integral of 1/(x^2 + a^2)?

The integral of

\frac{1}{x^2 + a^2}

\frac{1}{a}\tan^{-1}(\frac{x}{a}) + C

Solution of the integral of 1/(x^2 + a^2)

Solution: While we could straightly use some integral techniques, the reader should remind itself that there is a function that if you take the derivative of that, then we get exactly the solution of

\frac{1}{x^2 + a^2}

. Indeed, we get the following from this article:

\begin{align*} \frac{d}{dx} \frac{1}{a}\tan^{-1}\bigg(\frac{x}{a}\bigg) = \frac{1}{x^2 + a^2}. \end{align*}

Now take the antiderivative of

\frac{1}{x^2 - a^2}

, then we get that the integral of

\frac{1}{x^2 + a^2}

\frac{1}{a}\tan^{-1}(\frac{x}{a}) + C

Conclusion

Therefore, the integral of

\frac{1}{x^2 + a^2}

\frac{1}{a}\tan^{-1}(\frac{x}{a}) + C

, or in mathematical notation:

\begin{align*} \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\bigg(\frac{x}{a}\bigg) + C. \end{align*}

The post Integral of 1/(x^2 + a^2) appeared first on Epsilonify.

Integral of x/(x^2 – a^2)

The Mathematician — Fri, 26 May 2023 13:00:52 +0000

What is the integral of x/(x^2 – a^2)

The integral of

\frac{x}{x^2 - a^2}

\frac{1}{2}\ln \lvert x^2 - a^2 \rvert + C

Solution of the integral of x/(x^2 – a^2)

Solution: Before we determine the integral of

\frac{x}{x^2 - a^2}

, lets us recall what we need to show exactly:

\begin{equation*} \int \frac{x}{x^2 - a^2} dx. \end{equation*}

We use the substitution method where

u = x^2 - a^2

such that we get the derivative

du = 2xdx \iff xdx = \frac{1}{2}du

. Now we will implement that in the integral above, and therefore, we get the desired solution:

\begin{align*} \int \frac{x}{x^2 - a^2} dx &= \int \frac{\frac{1}{2}du}{u} \\ &= \frac{1}{2} \int \frac{du}{u} \\ &= \frac{1}{2} \ln \lvert u \rvert + C \\ &= \frac{1}{2} \ln \lvert x^2 - a^2 \rvert + C \end{align*}

Conclusion

The detailed solution above shows us that the integral of

\frac{x}{x^2 - a^2}

\frac{1}{2}\ln \lvert x^2 - a^2 \rvert + C

The post Integral of x/(x^2 – a^2) appeared first on Epsilonify.

Integral of x/(x^2 + a^2)

The Mathematician — Mon, 22 May 2023 13:00:26 +0000

What is the integral of x/(x^2 + a^2)?

The integral of

\frac{x}{x^2 + a^2}

\frac{1}{2}\ln(x^2 + a^2) + C

Solution of the integral of x/(x^2 + a^2)

We want to determine the integral of

\frac{x}{x^2 + a^2}

, i.e.,

\begin{align*} \int \frac{x}{x^2 + a^2} dx. \end{align*}

We will use the substitution method. Let

u = x^2 + a^2

. Then

du = 2xdx \iff \frac{1}{2}du = xdx

. So we get the following integral:

\begin{align*} \int \frac{x}{x^2 + a^2} dx &= \int \frac{\frac{1}{2}du}{u} \\ &= \frac{1}{2} \int \frac{du}{u} \\ &= \frac{1}{2}\ln \lvert u \rvert + C \\ &= \frac{1}{2}\ln (x^2 + a^2) + C. \end{align*}

Conclusion

So, the integral of

\frac{x}{x^2 + a^2}

\frac{1}{2}\ln(x^2 + a^2) + C

The post Integral of x/(x^2 + a^2) appeared first on Epsilonify.

Derivative of square root cos(x)

The Mathematician — Thu, 18 May 2023 13:00:24 +0000

What is the derivative of square root cos(x)?

The derivative of square root

\cos(x)

\frac{-\sin(x)}{2\sqrt{\cos(x)}}

Solution of the derivative of square root cos(x)

Let

F(x) = f(g(x)) = \sqrt{\cos(x)}

, where

f(u) = \sqrt{u}

and

g(x) = \cos(x)

. To find the derivative of

\sqrt{\cos(x)}

, we need to apply the chain rule:

\begin{align*} F'(x) = f'(g(x))g'(x). \end{align*}

We know from here that

f'(u) = \frac{1}{2\sqrt{u}}

and here that

g'(x) = -\sin(x)

. Therefore, we get:

\begin{align*} f'(g(x)) = \frac{1}{2\sqrt{g(x)}} = \frac{1}{2\sqrt{\cos(x)}} \quad \text{and} \quad g'(x) = -\sin(x). \end{align*}

Finally, this gives us the desired result:

\begin{align*} F'(x) &= f'(g(x))g'(x) \\ &= \frac{1}{2\sqrt{g(x)}} \cdot (-\sin(x)) \\ &= \frac{-\sin(x)}{2\sqrt{\cos(x)}}. \end{align*}

Conclusion

Therefore, the derivative of square root

\sqrt{\cos(x)}

\frac{-\sin(x)}{2\sqrt{\cos(x)}}

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Proof of Quotient Rule using First Principle of Derivatives

The Mathematician — Sun, 14 May 2023 13:00:42 +0000

How to prove the quotient rule derivative using first principle of derivatives

Proving the quotient rule can be easily done if you know to apply one trick, see the following section.

Proof of quotient rule derivative using first principle of derivatives

Let

f

and

g

be functions that are differentialbe at

x

and

g(x) \neq 0

. Then we want to prove the following formula:

\begin{align*} \bigg(\frac{f}{g}\bigg)'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}, \end{align*}

with the first principle of derivatives.

\begin{align*} \bigg(\frac{f}{g}\bigg)'(x) &= \lim_{h \rightarrow 0} \frac{\frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}}{h} \\ &= \lim_{h \rightarrow 0} \frac{f(x+h)g(x) - f(x)g(x+h)}{g(x+h)g(x)h}. \\ \end{align*}

Now the trick we need to apply is

- f(x)g(x) + f(x)g(x) = 0

and we place that in the above numerator:

\begin{align*} & \lim_{h \rightarrow 0} \frac{f(x+h)g(x) - f(x)g(x+h)}{g(x+h)g(x)h} = \\ & \lim_{h \rightarrow 0} \frac{f(x+h)g(x) - f(x)g(x) + f(x)g(x) - f(x)g(x+h)}{g(x+h)g(x)h} = \\ & \lim_{h \rightarrow 0} \frac{g(x)(f(x+h) - f(x)) - f(x)(g(x + h) - g(x))}{g(x+h)g(x)h} = \\ & g(x)\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{g(x+h)g(x)h} - f(x)\lim_{h \rightarrow 0} \frac{g(x + h) - g(x)}{g(x+h)g(x)h} = \\ & \frac{g(x)f'(x)}{(g(x))^2} - \frac{f(x)g'(x)}{(g(x))^2} = \\ & \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}, \end{align*}

where we have now proved the quotient rule.

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Product Rule Derivative using First Principle of Derivatives

The Mathematician — Wed, 10 May 2023 13:00:08 +0000

How to prove the product rule derivative using first principle of derivatives

We will prove the product rule by the first principle of derivatives, the definition of the derivative. In other words, we will prove the next equality holds:

\begin{equation*} (fg)'(x) = f'(x)g(x) + f(x)g'(x). \end{equation*}

Proof of product rule derivative using first principle of derivatives

\begin{align*} (fg)'(x) &= \lim_{h \rightarrow 0} \frac{f(x + h)g(x + h) - f(x)g(x)}{h} \\ \end{align*}

We will implement in the numerator the next equality because it equals zero and therefore won’t change:

\begin{equation*} f(x)g(x + h) - f(x)g(x + h) = 0. \end{equation*}

So we get

\begin{align*} &\lim_{h \rightarrow 0} \frac{f(x + h)g(x + h) - f(x)g(x)}{h} = \\ &\lim_{h \rightarrow 0} \frac{f(x + h)g(x + h) - f(x)g(x + h) + f(x)g(x + h) - f(x)g(x)}{h} = \\ &\lim_{h \rightarrow 0} \frac{(f(x + h)-f(x))g(x + h) + f(x)(g(x + h) - g(x))}{h} = \\ & \lim_{h \rightarrow 0} \frac{f(x + h)-f(x)}{h}g(x + h) + f(x) \lim_{h \rightarrow 0} \frac{(g(x + h) - g(x))}{h} = \\ & f'(x)g(x) + f(x)g'(x) \end{align*}

because

\begin{equation*} \lim_{h \rightarrow 0} \frac{f(x + h)-f(x)}{h} = f'(x) \quad \text{and} \quad \lim_{h \rightarrow 0} f(x) \frac{(g(x + h) - g(x))}{h} = g'(x). \end{equation*}

Therefore

(fg)'(x) = f'(x)g(x) + f(x)g'(x)

, which completes the proof.

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Reciprocal Rule Derivative using First Principle of Derivatives

The Mathematician — Sat, 06 May 2023 13:00:32 +0000

How to prove reciprocal rule derivative using first principle of derivatives?

We will prove the reciprocal rule by the first principle method, the definition of the derivative. The Reciprocal Rule is defined as

\begin{equation*} \bigg(\frac{1}{f}\bigg)' (x) = \frac{-f'(x)}{(f(x))^2}. \end{equation*}

We will prove that this equality holds.

Proof of Reciprocal Rule Derivative using First Principle of Derivatives

\begin{align*} \bigg(\frac{1}{f}\bigg)' (x) &= \lim_{h \rightarrow 0} \frac{\frac{1}{f(x + h)} - \frac{1}{f(x)}}{h} \\ &= \lim_{h \rightarrow 0} \frac{\frac{f(x) - f(x + h)}{f(x + h)f(x)}}{h} \\ &= \lim_{h \rightarrow 0} \frac{(- f(x+h) + f(x))}{h} \cdot \lim_{h \rightarrow 0} \frac{1}{f(x + h)f(x)} \\ &= - \lim_{h \rightarrow 0} \frac{(f(x+h) - f(x))}{h} \cdot \lim_{h \rightarrow 0} \frac{1}{f(x + h)f(x)} \\ &= f'(x) \cdot \lim_{h \rightarrow 0} \frac{1}{f(x + h)f(x)} \\ &= f'(x) \cdot \frac{1}{f(x)f(x)} \\ &= \frac{-f'(x)}{(f(x))^2}. \end{align*}

This shows that the Reciprocal Rule indeed holds, which completes the proof.

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Derivative of sec^3(x)

The Mathematician — Tue, 02 May 2023 13:00:46 +0000

What is the derivative of $\sec^3(x)$ ?

The derivative of

\sec^3(x)

3\tan(x)\sec^3(x)

Solution of the derivative of $\sec^3(x)$ .

Let

F(x) = f(g(x)) = \sec^3(x)

, where

f(u) = u^3

and

g(x) = \sec(x)

. Then to determine the derivative of

\sec^3(x)

, we need to apply the chain rule:

\begin{align*} F'(x) = f'(g(x))g'(x). \end{align*}

It is easy to see that

f'(u) = 3u^2

and we have seen here that

g'(x) = \tan(x)\sec(x)

. So we get:

\begin{align*} f'(g(x)) = f'(\sec(x)) = 3\sec^2(x) \quad \text{and} \quad g'(x) = \tan(x)\sec(x). \end{align*}

Substituting everything, we get:

\begin{align*} F'(x) &= f'(g(x))g'(x) \\ &= 3\sec^2(x)\tan(x)\sec(x) \\ &= 3\tan(x)\sec^3(x). \end{align*}

Conclusion

So, the derivative of

\sec^3(x)

3\tan(x)\sec^3(x)

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Calculus Archives - Epsilonify

Derivative of sin^4(x)

What is the derivative of sin^4(x)?

Solution of the derivative of sin^4(x)

Integral of 1/(x^2 – a^2)

What is the integral of 1/(x^2 – a^2)?

Solution of the integral 1/(x^2 – a^2)

Conclusion

Integral of 1/(x^2 + a^2)

What is the integral of 1/(x^2 + a^2)?

Solution of the integral of 1/(x^2 + a^2)

Conclusion

Integral of x/(x^2 – a^2)

What is the integral of x/(x^2 – a^2)

Solution of the integral of x/(x^2 – a^2)

Conclusion

Integral of x/(x^2 + a^2)

What is the integral of x/(x^2 + a^2)?

Solution of the integral of x/(x^2 + a^2)

Conclusion

Derivative of square root cos(x)

What is the derivative of square root cos(x)?

Solution of the derivative of square root cos(x)

Conclusion

Proof of Quotient Rule using First Principle of Derivatives

How to prove the quotient rule derivative using first principle of derivatives

Proof of quotient rule derivative using first principle of derivatives

Product Rule Derivative using First Principle of Derivatives

How to prove the product rule derivative using first principle of derivatives

Proof of product rule derivative using first principle of derivatives

Reciprocal Rule Derivative using First Principle of Derivatives

How to prove reciprocal rule derivative using first principle of derivatives?

Proof of Reciprocal Rule Derivative using First Principle of Derivatives

Derivative of sec^3(x)

What is the derivative of \sec^3(x)?

Solution of the derivative of \sec^3(x).

Conclusion

What is the derivative of $\sec^3(x)$ ?

Solution of the derivative of $\sec^3(x)$ .