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What is the derivative of f of g of h of x using the chain rule?
Wiki User
∙ 11y ago
f'(g(h(x)))*g'(h(x))*h'(x) where the prime denote a derivative with respect to x.
Wiki User
∙ 11y ago
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How do you differentiate sine x squared?
Using the Chain Rule :derivative of (sinx)2 = 2(sinx)1 * (derivative of sinx)d/dx (Sinx)2 = 2(sinx)1 * [d/dx (Sinx)]d/dx (Sinx)2 = 2(sinx) * (cosx)d/dx (Sinx)2 = 2 (sinx) * (cosx)d/dx (Sinx)2 = 2 sin(x) * cos(x)
What does the derivative graph mean?
I am assuming the you are talking about the graph of the derivative. The graph of the derivative of F(x) is the graph such that, for any x, the value of x on the graph of the derivative of F(x) is the slope at point x in F(x).
What is the derivative of 2x?
-2
What is the derivative of the square root of cos x?
To find the derivatve of the square root of cos x: Use the chain rule; this means multiply the inner derivative by the outer derivative. You can write the question f(x) = (cos x)1/2 This general break-down explains how to find d/dx f(x) note: d/dx basically symbolizes "the derivative of" In general terms: f(x) = x1/2 g(x) = cos x f(g(x)) = (cos x)1/2 outer derivative: d/dx f(z) = (1/2)*x-1/2 = 1/(4cos x)1/2 inner derivative: d/dx g(x) = -sin(x) final answer: d/dx f(g(x)) = -sin(x)/(4*cos x)1/2 note: d/dx means "the derivative of"; so d/dx x = 1 Further explained: Set up the equation to a more general form: (cos x)1/2 To make the inner derivative, look at cos(x) To make the outer derivative, look at x1/2 note: x ~ cos x; so we treat (cos x) simply as x, to create the outer derivative You probably know the necessary derivates: 1. derivative of cos x = -sin x 2. derivative of a1/2 = (1/2)*a-1/2 = 1/(4a)1/2 Multiplying the two we get the answer: -sin(x)/(4cos x)1/2
How do you do integration math?
Well if you are talking about calculus, integration is the anti-derivative. So as my teacher explained to us, instead of going down, you will go up. For example if you have the F(x) = 2x, the F'(x) = 2. F'(x) is the derivative here, so you will do the anti of a derivative. So with the same F(x) = 2x the integral, is SF(x) = 1/3x^3. The Integral will find you the area under the curve.
What is the derivative of cos 2x?
f'(x)=-sin2x(2) f'(x)=-2sin2x First do the derivative of cos u, which is -sin u. Then because of the chain rule, you have to take the derivative of what's inside and the derivative of 2x is 2.
What is the first second and third derivative of ln 1-x?
Original function: f(x) = ln(1-x)Using the chain rule for logarithms: (ln(u))' = 1/u * u'The first derivative: f'(x) = 1/(1-x) * -1= 1/(-1*(1-x))= 1/(x-1)= (x-1)^(-1)Using the chain rule for polynomials: (au^n)' = nau^(n-1) * u'The second derivative: f''(x) = -1*(x-1)^-2 * 1= -(x-1)^-2= -1/((x-1)^2)Using the chain rule again:The third derivative: f'''(x) = 2*(x-1)^-3 * 1= 2(x-1)^-3= 2/((x-1)^3)
What is the derivative of pi squared raised to the x?
f(x)=(pi2)x=pi2x. The derivative of kx=ln(k)*kx, so f'(x)=2ln(pi)*pi2x (with chain rule).
What is the Derivative chain rule of (4-x)^3?
The chain rule states that the derivative of f(g(x)) is f'(g(x))⋅g'(x). In other words, it helps us differentiate *composite functions*. For example, sin(x²) is a composite function because it can be constructed as f(g(x)) for f(x)=sin(x) and g(x)=x².
How do I calculate the first derivative of a function?
To calculate the first derivative of a function, you can follow these general steps: Identify the function: Determine the function for which you want to find the first derivative. Let's assume your function is denoted as f(x). Express the function: Write down the function in its general form, considering any constants or variables involved. For example, f(x) = 3x^2 + 2x - 1. Differentiate the function: Use differentiation rules to find the derivative of the function. The derivative represents the rate of change of the function with respect to the variable. For example, to differentiate f(x) = 3x^2 + 2x - 1, apply the power rule and the sum rule as follows: Power rule: For a term of the form ax^n, the derivative is d/dx(ax^n) = anx^(n-1). Sum rule: The derivative of a sum of functions is the sum of their derivatives. Applying these rules to the function f(x) = 3x^2 + 2x - 1: The derivative of the term 3x^2 is 6x (using the power rule). The derivative of the term 2x is 2 (using the power rule, where the exponent is 1). The derivative of the constant term -1 is 0 (as the derivative of a constant is always 0). So, the first derivative of f(x) = 3x^2 + 2x - 1 is f'(x) = 6x + 2. Simplify if necessary: If there are any further simplifications or rearrangements possible, apply them to obtain the final form of the first derivative. In summary, the process involves differentiating each term of the function with respect to the variable and then simplifying the resulting expression. Differentiation rules such as the power rule, sum rule, product rule, and chain rule can be used depending on the complexity of the function.
What is the derivative of 1 divided by sinx cosx?
use the chain rule and product rule: f(x)=1/u(x), f'(x)=-1/u2(x)*u'(x) and f(x)=a(x)b(x), f'(x)=a'(x)b(x)+b'(x)a(x) the derivative of sin is cos, and the derivative of cos in negative sin. so f(x)=1/(sin(x)cos(x)), f'(x)=-1/(sin(x)cos(x))2*(cos2(x)-sin2(x)) or assume sin(x)cos(x)=.5sin(2x) so using chain rule, f'(x)=-2/sin2(2x)*.5cos(2x)*2. Both answers are equivalent.
What is the derivative of ln 1-x?
In this case, you need to apply the chain rule. Note that the derivative of ln N = 1/N. In that case we get: f(x) = ln(1 - x) ∴ f'(x) = 1/(1 - x) × -1 ∴ f'(x) = -1/(1 - x)
How do you find the derivative of f of x equals 6 over x plus 3?
The original equation f(x) = 6/(x+3) can be rewritten as f(x) = 6(x+3)-1. Now derive the equation according the the power rule and the chain rule: y = 6 (x+3)-1 dy/dx = 6 (-1)(x+3)-2(1)* dy/dx = -6/(x+3)2 * by the chain rule, you must multiply by the derivative of which is simply one. Thus, the derivative of f(x) = 6/(x+3) equals -6/(x+3)2
What does formula of derivative of an inverse allows you to do?
The formula for the derivative of an inverse (finv)' = 1/(f' o (finv)) allows you get a formula for the derivative of the inverse of any function that you already know the derivative of. For example: What is the derivative of sqrt(x)? You could figure this out using the definition of the derivative, but it is complicated. You already know that the derivative of x2 is 2x. So let f = x2; finv = sqrt(x), f' = 2x. This gives: (sqrt(x))' = 1/(2 sqrt(x)). Now you have derived a "square root rule" with almost no work.
How do you differentiate sine x squared?
Using the Chain Rule :derivative of (sinx)2 = 2(sinx)1 * (derivative of sinx)d/dx (Sinx)2 = 2(sinx)1 * [d/dx (Sinx)]d/dx (Sinx)2 = 2(sinx) * (cosx)d/dx (Sinx)2 = 2 (sinx) * (cosx)d/dx (Sinx)2 = 2 sin(x) * cos(x)
What describes the product rule?
The product rule for derivatives is as follows. For the derivative of the product of two functions, "f" and "g":(f times g)' = f times g' + f' times g
What is the derivative of xlnx?
ln(x)+1. This can be found using the product rule: if f(x)=uv, f'(x)=u'v+v'u, so if u is x and v is ln(x), the product rule gives (1)(ln(x))+(x)(1/x)=ln(x)+1.
