First Derivative Calculator(Solver) with Steps
Free derivatives calculator(solver) that gets the detailed solution of the first derivative of a function.
-
Let [tex]u = - 3 x \sin^{2}{\left (x \right )} + 1[/tex].
-
The derivative of [tex]e^{u}[/tex] is itself.
-
Then, apply the chain rule. Multiply by [tex]\frac{d}{d x}\left(- 3 x \sin^{2}{\left (x \right )} + 1\right)[/tex]:
-
Differentiate [tex]- 3 x \sin^{2}{\left (x \right )} + 1[/tex] term by term:
-
The derivative of the constant [tex]1[/tex] is zero.
-
The derivative of a constant times a function is the constant times the derivative of the function.
-
Apply the product rule:
[tex]\frac{d}{d x}\left(f{\left (x \right )} g{\left (x \right )}\right) = f{\left (x \right )} \frac{d}{d x} g{\left (x \right )} + g{\left (x \right )} \frac{d}{d x} f{\left (x \right )}[/tex]
[tex]f{\left (x \right )} = x[/tex]; to find [tex]\frac{d}{d x} f{\left (x \right )}[/tex]:
-
Apply the power rule: [tex]x[/tex] goes to [tex]1[/tex]
[tex]g{\left (x \right )} = \sin^{2}{\left (x \right )}[/tex]; to find [tex]\frac{d}{d x} g{\left (x \right )}[/tex]:
-
Let [tex]u = \sin{\left (x \right )}[/tex].
-
Apply the power rule: [tex]u^{2}[/tex] goes to [tex]2 u[/tex]
-
-
Then, apply the chain rule. Multiply by [tex]\frac{d}{d x} \sin{\left (x \right )}[/tex]:
-
The derivative of sine is cosine:
[tex]\frac{d}{d x} \sin{\left (x \right )} = \cos{\left (x \right )}[/tex]
The result of the chain rule is:
[tex]2 \sin{\left (x \right )} \cos{\left (x \right )}[/tex]
-
The result is: [tex]2 x \sin{\left (x \right )} \cos{\left (x \right )} + \sin^{2}{\left (x \right )}[/tex]
-
So, the result is: [tex]- 6 x \sin{\left (x \right )} \cos{\left (x \right )} - 3 \sin^{2}{\left (x \right )}[/tex]
-
The result is: [tex]- 6 x \sin{\left (x \right )} \cos{\left (x \right )} - 3 \sin^{2}{\left (x \right )}[/tex]
-
The result of the chain rule is:
[tex]\left(- 6 x \sin{\left (x \right )} \cos{\left (x \right )} - 3 \sin^{2}{\left (x \right )}\right) e^{- 3 x \sin^{2}{\left (x \right )} + 1}[/tex]
Now simplify:
[tex]- \frac{3}{2} \left(2 x \sin{\left (2 x \right )} - \cos{\left (2 x \right )} + 1\right) e^{\frac{3 x}{2} \left(\cos{\left (2 x \right )} - 1\right) + 1}[/tex]
The answer is:
[tex]- \frac{3}{2} \left(2 x \sin{\left (2 x \right )} - \cos{\left (2 x \right )} + 1\right) e^{\frac{3 x}{2} \left(\cos{\left (2 x \right )} - 1\right) + 1}[/tex]
* is multiplication
oo is $\infty$
pi is $\pi$
x^2 is x2
sqrt(x) is $\sqrt{x}$
sqrt[3](x) is $\sqrt[3]{x}$
(a+b)/(c+d) is $\frac{a+b}{c+d}$
The Most Important Derivatives - Basic Formulas/Rules
$\frac{d}{dx}a=0$ (a is a constant)
$\frac{d}{dx}x=1$
$\frac{d}{dx}x^n=nx^{n-1}$
$\frac{d}{dx}e^x=e^x$
$\frac{d}{dx}\log x=\frac1x$
$\frac{d}{dx}a^x=a^x\log x$
$(f\ g)' = f'g + fg'$ - Product Rule
$(\frac{f}{g})' = \frac{f'g - fg'}{g^2}$ - Quotient Rule
$\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$ - Chain Rule
$\frac{d}{dx}\sin(x)=\cos(x)$
$\frac{d}{dx}\cos(x)=-\sin(x)$
$\frac{d}{dx}\tan(x)=\sec^2(x)$
$\frac{d}{dx}\cot(x)=-csc^2(x)$
$\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}\arccos(x)=-\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}$
$\frac{d}{dx}\text{arccot}(x)=-\frac{1}{1+x^2}$
$\frac{d}{dx}\text{arcsec}(x)=\frac{1}{x\sqrt{x^2-1}}$
$\frac{d}{dx}\text{arccsc}(x)=-\frac{1}{x\sqrt{x^2-1}}$