
Find \[\dfrac{{dy}}{{dx}}\], when \[y = \sin \left( {{x^x}} \right).\]
Answer
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Hint: Here, we have to find the derivative of the function with respect to the variable \[x\]. 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: \[\dfrac{d}{{dx}}(\sin x) = \cos x\]; \[\dfrac{d}{{dx}}(\log x) = \dfrac{1}{x}\];
2.Logarithmic formula: \[\log {a^b} = b\log a\]
3.Product rule: \[\dfrac{d}{{dx}}(uv) = uv' + vu'\] where \[u',v'\] are the derivatives of \[u,v\] respectively.
Complete step-by-step answer:
We have to find the derivative of \[y\] with respect to \[x\].
\[y = \sin ({x^x})\]
Taking derivative on both the sides, we have
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\sin ({x^x})} \right)\]
By using differential formula\[\dfrac{d}{{dx}}(\sin x) = \cos x\], we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \cos ({x^x})\dfrac{d}{{dx}}({x^x})\] …………………………..(1)
Now let us take \[{x^x}\] as \[p\].
\[p = {x^x}\]
Taking logarithms on both the sides, we get
\[ \Rightarrow \log p = \log {x^x}\]
By using Logarithmic formula\[\log {a^b} = b\log a\] , we have
\[ \Rightarrow \log p = x\log x\]
By using product rule \[\dfrac{d}{{dx}}(uv) = uv' + vu'\], we get
\[ \Rightarrow \dfrac{1}{p}\dfrac{{dp}}{{dx}} = x \cdot \dfrac{1}{x} + \log x \cdot 1\]
By multiplication, we have
\[ \Rightarrow \dfrac{1}{p}\dfrac{{dp}}{{dx}} = 1 + \log x\]
Multiplying by \[p\] on both the sides, we get
\[ \Rightarrow \dfrac{{dp}}{{dx}} = p(1 + \log x)\]
Again substituting \[p\] as \[{x^x}\], we have
\[ \Rightarrow \dfrac{{d({x^x})}}{{dx}} = {x^x}(1 + \log x)\]
Substituting the derivative of \[{x^x}\] in equation (1), we get
\[\dfrac{{dy}}{{dx}} = {x^x}\cos ({x^x})(1 + \log x)\]
Therefore \[\dfrac{{dy}}{{dx}} = {x^x}\cos ({x^x})(1 + \log x)\]
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.
Formula Used:
We will use the following formulas:
1.Differential formula: \[\dfrac{d}{{dx}}(\sin x) = \cos x\]; \[\dfrac{d}{{dx}}(\log x) = \dfrac{1}{x}\];
2.Logarithmic formula: \[\log {a^b} = b\log a\]
3.Product rule: \[\dfrac{d}{{dx}}(uv) = uv' + vu'\] where \[u',v'\] are the derivatives of \[u,v\] respectively.
Complete step-by-step answer:
We have to find the derivative of \[y\] with respect to \[x\].
\[y = \sin ({x^x})\]
Taking derivative on both the sides, we have
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\sin ({x^x})} \right)\]
By using differential formula\[\dfrac{d}{{dx}}(\sin x) = \cos x\], we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \cos ({x^x})\dfrac{d}{{dx}}({x^x})\] …………………………..(1)
Now let us take \[{x^x}\] as \[p\].
\[p = {x^x}\]
Taking logarithms on both the sides, we get
\[ \Rightarrow \log p = \log {x^x}\]
By using Logarithmic formula\[\log {a^b} = b\log a\] , we have
\[ \Rightarrow \log p = x\log x\]
By using product rule \[\dfrac{d}{{dx}}(uv) = uv' + vu'\], we get
\[ \Rightarrow \dfrac{1}{p}\dfrac{{dp}}{{dx}} = x \cdot \dfrac{1}{x} + \log x \cdot 1\]
By multiplication, we have
\[ \Rightarrow \dfrac{1}{p}\dfrac{{dp}}{{dx}} = 1 + \log x\]
Multiplying by \[p\] on both the sides, we get
\[ \Rightarrow \dfrac{{dp}}{{dx}} = p(1 + \log x)\]
Again substituting \[p\] as \[{x^x}\], we have
\[ \Rightarrow \dfrac{{d({x^x})}}{{dx}} = {x^x}(1 + \log x)\]
Substituting the derivative of \[{x^x}\] in equation (1), we get
\[\dfrac{{dy}}{{dx}} = {x^x}\cos ({x^x})(1 + \log x)\]
Therefore \[\dfrac{{dy}}{{dx}} = {x^x}\cos ({x^x})(1 + \log x)\]
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.
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