List of Derivative Rules | All Derivative Rules

You can find all the derivative rules below. We wish everyone good work and good lessons...

Below is a list of all the derivative rules.

Constant Rule:

$f(x)=c$ then $f'(x)=0$

Constant Multiple Rule:

$g(x)=c.f(x)$ then $g'(x)=c.f'(x)$

Power Rule:

$f(x)=x^n$ then $f'(x)=nx^{n-1}$

Sum and Difference Rule:

$h(x)=f(x) ± g(x)$ then $h'(x)=f'(x) ± g'(x)$

Product Rule:

$h(x)=f(x).g(x)$ then $h'(x)=f'(x).g(x)+f(x).g'(x)$

Quotient Rule:

$h(x)=\frac{f(x)}{g(x)}$ then $h'(x)=\frac{f'(x).g(x)-f(x).g'(x)}{g(x)^2}$

Chain Rule:

$h(x)=f(g(x))$ then $h'(x)=f'(g(x)).g'(x)$

Trig Derivatives:

$f(x)=sin(x)$ then $f'(x)=cos(x)$

$f(x)=cos(x)$ then $f'(x)=-sin(x)$

$f(x)=tan(x)$ then $f'(x)=sec^2(x)$

$f(x)=sec(x)$ then $f'(x)=sec(x).tan(x)$

$f(x)=cot(x)$ then $f'(x)=-csc^2(x)$

$f(x)=csc(x)$ then $f'(x)=-csc(x).cot(x)$

Exponential Derivatives:

$f(x)=a^x$ then $f'(x)=ln(a).a^x$

$f(x)=e^x$ then $f'(x)=e^x$

$f(x)=a^{g(x)}$ then $f'(x)=ln(a).a^{g(x)}.g'(x)$

$f(x)=e^{g(x)}$ then $f'(x)=e^{g(x)}.g'(x)$

Logarithm Derivatives:

$f(x)=log_a(x)$ then $f'(x)=\frac{1}{ln(a).x}$

$f(x)=lnx$ then $f'(x)=\frac{1}{x}$

$f(x)=log_a(g(x))$ then $f'(x)=\frac{g'(x)}{ln(a).g(x)}$

$f(x)=ln(g(x))$ then $f'(x)=\frac{g'(x)}{g(x)}$

Read More

arccot(x) integral | What is integrate of arccotx or cot^-1x?

Greetings dear friends. In this article, we will share with you what is the integrate of arccot(x).

arccot integrate

Integral of arccot(x) = $x.arccot(x)+\frac{1}{2}.ln(1+x^2)+C$

>>> $arccot(x)=cot^{-1}x$

>>> $\int arccot(x).dx=\int cot^{-1}x.dx$

>>> $\int arccot(x).dx=x.arccot(x)+\frac{1}{2}.ln(1+x^2)+C$

Read More

arctan(x) integral | What is integrate of arctanx or tan^-1x?

Greetings dear friends. In this article, we will share with you what is the integrate of arctan(x).

arctan integrate

Integral of arctan(x) = $x.arctan(x)-\frac{1}{2}.ln(1+x^2)+C$

>>> $arctan(x)=tan^{-1}x$

>>> $\int arctan(x).dx=\int tan^{-1}x.dx$

>>> $\int arctan(x).dx=x.arctan(x)-\frac{1}{2}.ln(1+x^2)+C$

Read More

arccos(x) integral | What is integrate of arccosx or cos^-1x?

 Greetings dear friends. In this article, we will share with you what is the integrate of arccos(x).

arccos integrate

Integral of arccos(x) = $x.arccos(x)-\sqrt{1-x^2}+C$

>>> $arccos(x)=cos^{-1}x$

>>> $\int arccos(x).dx=\int cos^{-1}x.dx$

>>> $\int arccos(x).dx=x.arccos(x)-\sqrt{1-x^2}+C$

Read More

arcsin(x) integral | What is integrate of arcsinx or sin^-1x?

Greetings dear friends. In this article, we will share with you what is the integrate of arcsin(x).

arcsin integrate

Integral of arcsin(x) = $x.arcsin(x)+\sqrt{1-x^2}+C$

>>> $arcsin(x)=sin^{-1}x$

>>> $\int arcsin(x).dx=\int sin^{-1}x.dx$

>>> $\int arcsin(x).dx=x.arcsin(x)+\sqrt{1-x^2}+C$

Read More

$\int \frac{x}{x-1}dx$ integral | What is the integrate of x/x-1 ?

Hello everyone. This lesson we will tell the integrate of $\int \frac{x}{x-1}dx$.

What is the integral of x/x-1?

integral of x/x-1

First we separate the expression.

So;

$\frac{x}{x-1}=1+\frac{1}{x-1}$

So our answer;

$\int \frac{x}{x-1}dx=\int 1 dx+ \int \frac{1}{x-1}dx=x+ln|x-1|+C$

Read More

$\sqrt{1+\sqrt{1+\sqrt{x}}}$ derivative | What is the derivative of $\frac{d}{dx}\sqrt{1+\sqrt{1+\sqrt{x}}}$?

Hello everyone. This lesson we will tell the derivative of $\sqrt{1+\sqrt{1+\sqrt{x}}}$.

What is the derivative of $\sqrt{1+\sqrt{1+\sqrt{x}}}$?

Compute the derivative of $\sqrt{1+\sqrt{1+\sqrt{x}}}$.

Here we have a more complicated chain of compositions, so we use the chain rule twice. At the outermost “layer” we have the function $g(x)=1+\sqrt{1+\sqrt{x}}$ plugged into $f(x)=\sqrt{x}$, so applying the chain rule once gives

$\frac{d}{dx}\sqrt{1+\sqrt{1+\sqrt{x}}}=\frac{1}{2}(1+\sqrt{1+\sqrt{x}})^{\frac{-1}{2}}.\frac{d}{dx}(1+\sqrt{1+\sqrt{x}})$

Not we need the derivative of $\sqrt{1+\sqrt{x}}$. Using the chain rule again:

$\frac{d}{dx}\sqrt{1+\sqrt{x}}=\frac{1}{2}(1+\sqrt{x})^{\frac{-1}{2}}.\frac{1}{2}x^{\frac{-1}{2}}$

So the original derivative is

$\frac{d}{dx}\sqrt{1+\sqrt{1+\sqrt{x}}}=\frac{1}{2}(1+\sqrt{1+\sqrt{x}})^{\frac{-1}{2}}.\frac{1}{2}(1+\sqrt{x})^{\frac{-1}{2}}.\frac{1}{2}x^{\frac{-1}{2}}$

=$\frac{1}{8\sqrt{x}.\sqrt{1+\sqrt{x}}.\sqrt{1+\sqrt{1+\sqrt{x}}}}$

Using the chain rule, the power rule, and the product rule, it is possible to avıid using the quotient rule entirely.

Read More