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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)
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What is the differentiation of y with respect to x for x equals y exponent y?
x = yy differentiate both sides with respect to x dx = (y * yy-1) dy dy/dx = y * yy-1 dy/dx = yy = x hence differentiate of y wrt x is x only
When are y equals sin x and y equals cos x equal?
y=sinx y=cosxsinx=cosx=>sinx/cosx=1=>tanx=1=>x=45oie.. y=sin45=cos45y=1/(square root of 2)
Arcsin x equals sin-1?
yes y=sinx is x=arcsiny
How do you differentiate xy equals logy c?
The differential of the product xy with respect to x is y + x dy/dx. The differential of logy with respect to x is (1/y) dy/dx. The role of c in this question is not made clear.
Differentiate y equals a power x?
y=ax y'=ln(a)*ax
What is the derivative of sin x to the e to the xth power?
y = (sinx)^(e^x) ln(y) = ln((sinx)^(e^x)) ln(y) = (e^x)ln(sinx) (1/y)dy = (e^x)(1/sinx)(cosx)+ln(sinx)(e^x)dx (1/y)dy = (e^x)(cotx)+ln(sinx)(e^x)dx dy = ((sinx)^(e^x))((cotx)(e^x)+ln(sinx)(e^x))dx dy = ((e^x)(sinx)^(e^x))(cotx+ln(sinx))dx
What is the differentiation of y with respect to x for x equals y exponent y?
x = yy differentiate both sides with respect to x dx = (y * yy-1) dy dy/dx = y * yy-1 dy/dx = yy = x hence differentiate of y wrt x is x only
What is the transformation that maps y equals sinx onto y equals the inverse of sinx?
f(x) = 1/x except where x = 0.
When are y equals sin x and y equals cos x equal?
y=sinx y=cosxsinx=cosx=>sinx/cosx=1=>tanx=1=>x=45oie.. y=sin45=cos45y=1/(square root of 2)
C program to find the value of sinx?
#include double x, y;y = sin (x);
How do you differentiate xy equals logy c?
The differential of the product xy with respect to x is y + x dy/dx. The differential of logy with respect to x is (1/y) dy/dx. The role of c in this question is not made clear.
Arcsin x equals sin-1?
yes y=sinx is x=arcsiny
What is the derivative of 1 divided by sinx?
y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx
What is sin 3x in terms of sin x?
given the identity sin(x+y)=sinx cosy + siny cosxsin2x = 2 sinx cosx andsin(2(x)+x) = sin 2x cos x + sinx cos 2xusing the last two identities givessin3x= 2 sinx cosx cosx + sinx cos2xfactoring the sinx we havesin3x = sinx(2cosx cosx+cos2x)which satisfies the requirement.However, we can simplify further since cos 2x = cosx cosx - sinx sinx (a well known identity)sin3x = sinx (2cosx cosx +cosx cosx - sinx sinx)so sin3x= sinx(3cosx cosx - sinx sinx)or sin 3x = 3.cos²x.sinx - sin³x* * * * *Good, but not good enough. The answer was required in terms of sin, not a mixture of sinx and cosx. Easily recitified, though, since cos²x = 1 - sin²xTherefore sin3x = 3*(1-sin²x)*sinx - sin³x= 3sinx - 3sin³x - sin³x= 3sinx - 4sin³x
Differentiate with respect to x xy-y3 equals 1?
Differentiating with respect to one variable means thinking of all the other variables as constants. The answer is thus y=0.
How do you differentiate sin sin x?
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))
Differentiate 2 to the power of x with respect to x?
so say y = 2^x dy/dx = ln2.2^x (. = multiplication symbol, ^ = to-the-power-of symbol) The general formula is (where 'a' is a constant, x is what you are differentiating with respect to and y is f(x)) y = a^x then dy/dx = lna.a^x Go ask a math teacher or look up exponential function differentiation on the internet for why.
