
Differentiate the expression \[{{x}^{{{x}^{2}}}}\] with respect to x.
Answer
596.7k+ views
Hint: We will first assume y equal to \[{{x}^{{{x}^{2}}}}\] and then take log on both sides and after simplifying we will differentiate it implicitly on left hand side and on the right hand side we will use product rule \[\dfrac{d(uv)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\] to differentiate.
Complete step-by-step answer:
Before proceeding with the question. We should understand differentiation and concepts related to it.
Differentiation means ratio of rate of change between two variables. In which the variable in the numerator is dependent on the variable in the denominator. Implicit differentiation is an approach to taking derivatives that uses the chain rule to avoid solving explicitly for one of the variables. For example, if \[y+3x=8\], we can directly take the derivative of each term with respect to x to obtain \[\dfrac{dy}{dx}+3=0\] so \[\dfrac{dy}{dx}=-3\]. The product rule of differentiation is \[\dfrac{d(uv)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\].
Let’s consider \[y={{x}^{{{x}^{2}}}}......(1)\]
Now we will take natural log on both sides of the equation (1), so we get,
\[\log y=\log {{x}^{{{x}^{2}}}}......(2)\]
Log of power rule is \[\log {{x}^{y}}=y\log x\]. Now applying this rule on the right hand side in equation (2) we get,
\[\log y={{x}^{2}}\log x......(3)\]
We will differentiate both sides in equation (3), on the left hand side we need to differentiate with respect to x so it is an implicit differentiation and on the right hand side we will use the product rule of differentiation.
Differentiation of log(x) is $\dfrac{d(log x)}{dx} = \dfrac{1}{x}$ and differentiation of ${x}^{2} $ is ${\dfrac {d{{x}^{2}}} {dx}= 2x} $
\[\dfrac{1}{y}\dfrac{dy}{dx}={{x}^{2}}\times \dfrac{1}{x}+2x\log x.......(4)\]
Now rearranging and simplifying equation (4) we get,
\[\dfrac{dy}{dx}=y\left( x+2x\log x \right).......(5)\]
Now substituting \[y={{x}^{{{x}^{2}}}}\] in equation (5) we get,
\[\dfrac{dy}{dx}={{x}^{{{x}^{2}}}}\left( x+2x\log x \right)\]
Hence the differentiation of the expression \[{{x}^{{{x}^{2}}}}\] with respect to x is \[{{x}^{{{x}^{2}}}}\left( x+2x\log x \right)\].
Note: Remembering basic differentiation formulas and logarithmic rules is the key here. Also we need to substitute for y in equation (5) to get the final answer. We can make a mistake in applying the log of power rule in equation (2) and we may also get confused in doing the implicit differentiation of the left hand side of equation (3).
Complete step-by-step answer:
Before proceeding with the question. We should understand differentiation and concepts related to it.
Differentiation means ratio of rate of change between two variables. In which the variable in the numerator is dependent on the variable in the denominator. Implicit differentiation is an approach to taking derivatives that uses the chain rule to avoid solving explicitly for one of the variables. For example, if \[y+3x=8\], we can directly take the derivative of each term with respect to x to obtain \[\dfrac{dy}{dx}+3=0\] so \[\dfrac{dy}{dx}=-3\]. The product rule of differentiation is \[\dfrac{d(uv)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\].
Let’s consider \[y={{x}^{{{x}^{2}}}}......(1)\]
Now we will take natural log on both sides of the equation (1), so we get,
\[\log y=\log {{x}^{{{x}^{2}}}}......(2)\]
Log of power rule is \[\log {{x}^{y}}=y\log x\]. Now applying this rule on the right hand side in equation (2) we get,
\[\log y={{x}^{2}}\log x......(3)\]
We will differentiate both sides in equation (3), on the left hand side we need to differentiate with respect to x so it is an implicit differentiation and on the right hand side we will use the product rule of differentiation.
Differentiation of log(x) is $\dfrac{d(log x)}{dx} = \dfrac{1}{x}$ and differentiation of ${x}^{2} $ is ${\dfrac {d{{x}^{2}}} {dx}= 2x} $
\[\dfrac{1}{y}\dfrac{dy}{dx}={{x}^{2}}\times \dfrac{1}{x}+2x\log x.......(4)\]
Now rearranging and simplifying equation (4) we get,
\[\dfrac{dy}{dx}=y\left( x+2x\log x \right).......(5)\]
Now substituting \[y={{x}^{{{x}^{2}}}}\] in equation (5) we get,
\[\dfrac{dy}{dx}={{x}^{{{x}^{2}}}}\left( x+2x\log x \right)\]
Hence the differentiation of the expression \[{{x}^{{{x}^{2}}}}\] with respect to x is \[{{x}^{{{x}^{2}}}}\left( x+2x\log x \right)\].
Note: Remembering basic differentiation formulas and logarithmic rules is the key here. Also we need to substitute for y in equation (5) to get the final answer. We can make a mistake in applying the log of power rule in equation (2) and we may also get confused in doing the implicit differentiation of the left hand side of equation (3).
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