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To differentiate y=sin(sin(x)) you need to use the chain rule.
A common way to remember the chain rule is "derivative of the outside, keep the inside, derivative of the inside".
First, you take the derivative of the outside. The derivative of sin is cos.
Then, you keep the inside, so you keep sin(x).
Then, you multiple by the derivative of the inside. Again, the derivative of sinx is cosx.
In the end, you get y'=cos(sin(x))cos(x))
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How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
What is sin x times sin x?
We write sin x * sin x = sin2 x
What is sin x differentiated?
The differential of sin x with respect to x is: d(sin x) = cos x dx
Why does the derivative of sin x equal - cos x?
It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.
How do you factor sin squared times x plus cos2x -cosx equals 0?
2
Differentiate sin x with respect to x?
cos x
How do you differentiate sin squared x plus cos squared x?
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
How do you differentiate x sinx with respect to x?
You should apply the chain rule d/dx(x.sin x) = x * d/dx(sin x) + sin x * d/dx(x) = x * cos x + sin x * 1 = x.cos x + sin x
Differentiate y equals xsinx plus cosx?
y = x sin(x) + cos(x)Derivative of the first term = x cos(x) + sin(x)Derivative of the second term = -sin(x)y' = Sum of the derivatives = x cos(x) + sin(x) - sin(x)= [ x cos(x) ]
How do you differentiate x sinx with respect to y?
If you actually mean "... with respect to x", and that y is equal to this function of x, then the answer is:y = x sin(x)∴ dy/dx = sin(x) + x cos(x)
How do you differentiate cosine square root of x?
The derivative of cos x is -sin x, the derivative of square root of x is 1/(2 root(x)). Applying the chain rule, the derivative of cos root(x) is -sin x times 1/(2 root(x)), or - sin x / (2 root x).
Can you differentiate the following function fx equals 9 times x to the 2nd power times sinxtanx?
f(x)=9x2(sin x * tan x)f'(x)= 18x(sin x * tan x) + 9x2(cos x * tan x + sec2x * sin x)there might be some identities that allow that to be simplified to look prettier
How you can defferentiate an integral?
If the upper limit is a function of x and the lower limit is a constant, you can differentiate an integral using the Fudamental Theorem of Calculus. For example you can integrate Integral of [1,x^2] sin(t) dt as: sin(x^2) d/dx (x^2) = sin(x^2) (2x) = 2x sin(x^2) The lower limit of integration is 1 ( a constant). The upper limit of integration is a function of x, here x^2. The function being integrated is sin(t)
How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
How do you differentiate a cosine function That is what is the derivative of the cosine of x with respect to the independent variable x?
The derivative of cosine of x is simply the negative sine of x. In mathematical terms f'(x) = d/dx[cos(x)] = -sin(x)
What is the derivative of 2cosx times the inverse of sinx?
d/dx (2cos(x)sin⁻¹(x)) right, 2cos(x)sin⁻¹(x) is a product, so we'll differentiate using the product rule: d/dx(uv) = u d/dx(v) + v d/dx(u) u = 2cos(x) v = sin⁻¹(x) d/dx(u) = -2sin(x) to find d/dx(sin⁻¹(x)) we'll set y=sin⁻¹(x) x=sin(y) dx/dy = cos(y) dy/dx = 1/(cos(y)) cos²(y) + sin²(y)=1, dy/dx = 1/sqrt(1-sin²y) = 1/sqrt(1-x²) [x=sin(y)] so plugging all this into our product rule, d/dx (2cos(x)sin⁻¹(x)) = 2cos(x)/sqrt(1-x²) - 2sin(x)sin⁻¹(x).
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.
