Enter the function and select the variable for which the tool will determine its derivative up to second order, with detailed calculations shown.
In mathematics, the second by-product is referred to as the second one-order by-product of the given expression. The locating method of the spinoff of an equation is known as the differentiation. therefore, the technique of determining the second one-order spinoff is referred to as 2d-order differentiation. If the feature is differentiated two instances, then we can get the second-order by-product of a certain expression.
In easy phrases, the second derivative calculates how the fee of alternate of a sure amount is converting itself. for instance, discover the second one by-product of the location of an item with appreciate to the price at which the rate of the item is converting with respect to time (t) is:
$$ m = d(v) / d(t) = d^2 x / dt^2 $$
in which,
m = acceleration
t = time
x = position
d = instantaneous change
v = velocity
When observe the power rule twice, it's going to create the second one derivative strength rule this is utilized by double by-product calculator as:
$$ d^2 / dx^2 [x^n] = d / dx . d / dx [x^n] = d / dx [nx^{x-1} ] = n (n – 1) x^{n-2} $$
the second derivative of a characteristic f(x) is typically denoted as:
F’’ = (f’)’
whilst using notation for derivatives, the second one spinoff of a structured variable (y) with respect the independent variable (x) is written as \( d^2y / dx^2 \)
That is derived from
$$ d^2y / dx^2 = d / dx (dy / dx) $$
Calculating the second by-product of any expression has emerge as reachable when you have proper understanding about strength and product guidelines.
Example:
Find the second derivative for \( d^2 / dx^2 sin (x) cos^3 (x) \).
Solution:
Given that:
$$ d^2 / dx^2 sin (x) cos^3 (x) $$
The second derivative calculator apply the product rule first:
$$ d / dx f(x) g(x) = f(x) d / dx g(x) + g(x) d / dx f(x) $$
$$ f(x)=cos^3(x); \text { to find } d / dx f(x): $$
$$ Let u=cos(x).$$
second spinoff take a look at calculator applies the electricity rule:
$$ U^3 \text{ goes to } 3u^2 $$
Then, 2nd order spinoff calculaor practice the chain rule. Multiply via d / dx cos(x):
The by-product of cosine is negative sine:
$$ d / dx cos(x) = −sin (x) $$
The end result of the chain rule is:
$$ −3 sin (x) cos^2 (x) $$
$$ g(x) = sin (x); to find d / dx g(x): $$
The derivative of sine is cosine:
$$ d / dx sin (x) = cos (x) $$
The result is: \( −3sin^2 (x) cos^2(x) + cos^4(x) \)
Now, double spinoff calculator simplify these obtained results:
$$ Cos (2x)^2 + cos (4x)^2 $$
Therefore, differentiate \( −3sin^2(x) cos^2(x) + cos^4(x) \) term by term:
Let u = cos (x)
Now, the second one derivative test calculator applies the strength rule:
\( u^4 \text{ goes to } 4u^3 \) Then, apply the chain rule. Multiply by \( d / dxcos(x) \):
The spinoff of cosine is poor sine:
$$ d / dx cos(x) = −sin(x) $$
The end result of the chain rule is:
$$ −4sin (x) cos^3 (x) $$
Now, 2nd derivative calculator apply the product rule once more for finding the second one by-product:
$$ d / dx f(x) g(x) = f(x) d / dx g (x) + g(x) d / dx f(x) $$
$$ f(x) = cos^2 (x); \text { to find } d / dx f(x): $$
Let u = cos(x).
Apply the power rule: \( 9 u^2 \text{ goes to } 2u \)
Then, apply the chain rule. Multiply by d / dx cos (x):
The derivative of cosine is negative sine: $$ d / dx cos (x) = −sin (x) $$
The result of the chain rule is:
$$ −2 sin (x) cos (x) $$
$$ g(x) = sin^2(x); \text{ to find} d / dx g(x): $$
Let u = sin (x).
Apply the power rule: u^2 \text{ goes to} 2u
Then, second order derivative calculator applies the chain rule. Multiply by d / dx sin (x):
The derivative of sine is cosine:
$$ d / dx sin (x) = cos(x) $$
The result of the chain rule is:
$$ 2 sin(x) cos(x) $$
So, the result is:
$$ −2sin^3 (x) cos (x) + 2sin (x) cos^3 (x) $$
Now, simplify :
$$ 6sin^3 (x) cos (x) − 6sin (x) cos^3 (x) $$
Hence,
$$ 6sin^3(x) cos(x) – 10 sin (x) cos^3 (x) $$
After simplifying the answer is:
$$ −sin (2x) − 2sin (4x) $$
The second one by-product is used to locate the local extrema of the feature underneath specific conditions. If a characteristic f has a vital point for which f&high;(x) = 0 and the second one by-product is positive (+ve) at this factor, then the function f has a local minimal here.
The signal of the second spinoff tells about its concavity. If the second one by-product is described on an interval (m, n) and f ''(x) > zero on a certain interval, then the derivative of the function is high-quality..
