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Question: Find \[\dfrac{{dy}}{{dx}}\], when \[y = \sin \left( {{x^x}} \right).\]...

Find dydx\dfrac{{dy}}{{dx}}, when y=sin(xx).y = \sin \left( {{x^x}} \right).

Explanation

Solution

Here, we have to find the derivative of the function with respect to the variable xx. First, we will find the derivative of the trigonometric function. Then we will find the derivative of the exponential function using the product rule. Differentiation can be defined as a derivative of a function with respect to an independent variable.

Formula Used:
We will use the following formulas:
1.Differential formula: ddx(sinx)=cosx\dfrac{d}{{dx}}(\sin x) = \cos x; ddx(logx)=1x\dfrac{d}{{dx}}(\log x) = \dfrac{1}{x};
2.Logarithmic formula: logab=bloga\log {a^b} = b\log a
3.Product rule: ddx(uv)=uv+vu\dfrac{d}{{dx}}(uv) = uv' + vu' where u,vu',v' are the derivatives of u,vu,v respectively.

Complete step-by-step answer:
We have to find the derivative of yy with respect to xx.
y=sin(xx)y = \sin ({x^x})
Taking derivative on both the sides, we have
dydx=ddx(sin(xx))\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\sin ({x^x})} \right)
By using differential formuladdx(sinx)=cosx\dfrac{d}{{dx}}(\sin x) = \cos x, we get
dydx=cos(xx)ddx(xx)\Rightarrow \dfrac{{dy}}{{dx}} = \cos ({x^x})\dfrac{d}{{dx}}({x^x}) …………………………..(1)
Now let us take xx{x^x} as pp.
p=xxp = {x^x}
Taking logarithms on both the sides, we get
logp=logxx\Rightarrow \log p = \log {x^x}
By using Logarithmic formulalogab=bloga\log {a^b} = b\log a , we have
logp=xlogx\Rightarrow \log p = x\log x
By using product rule ddx(uv)=uv+vu\dfrac{d}{{dx}}(uv) = uv' + vu', we get
1pdpdx=x1x+logx1\Rightarrow \dfrac{1}{p}\dfrac{{dp}}{{dx}} = x \cdot \dfrac{1}{x} + \log x \cdot 1
By multiplication, we have
1pdpdx=1+logx\Rightarrow \dfrac{1}{p}\dfrac{{dp}}{{dx}} = 1 + \log x
Multiplying by pp on both the sides, we get
dpdx=p(1+logx)\Rightarrow \dfrac{{dp}}{{dx}} = p(1 + \log x)
Again substituting pp as xx{x^x}, we have
d(xx)dx=xx(1+logx)\Rightarrow \dfrac{{d({x^x})}}{{dx}} = {x^x}(1 + \log x)
Substituting the derivative of xx{x^x} in equation (1), we get
dydx=xxcos(xx)(1+logx)\dfrac{{dy}}{{dx}} = {x^x}\cos ({x^x})(1 + \log x)
Therefore dydx=xxcos(xx)(1+logx)\dfrac{{dy}}{{dx}} = {x^x}\cos ({x^x})(1 + \log x)

Note: With the help of differentiation, we are able to find the rate of change of one quantity with respect to another. Some of the examples are acceleration, to calculate the highest and lowest point of the curve in a graph or to know its turning point, the derivative function is used and to find tangent and normal to a curve. The product rule states that if the two "parts" of the function are being multiplied together, and the chain rule is if they are being composed.