
What is the derivative of $y={{\pi }^{x}}+{{x}^{\pi }}?$
Answer
526.2k+ views
Hint: We use the basic differentiation formulae to solve this sum. We split the above question into two parts. We differentiate the two parts separately and add the two answers. For the first part we apply ln function and differentiate using the differentiation of ln formula. For the second part, we differentiate using the basic power rule of differentiation.
Complete step by step answer:
In order to solve this question, let us split the given question $y={{\pi }^{x}}+{{x}^{\pi }}$ into two parts ${{y}_{1}}$ and ${{y}_{2}}.$ Let ${{y}_{1}}={{\pi }^{x}}$ and ${{y}_{2}}={{x}^{\pi }},$ and we need to differentiate the two equations separately. Let us consider the first equation,
$\Rightarrow {{y}_{1}}={{\pi }^{x}}$
Let us apply ln on both sides now,
$\Rightarrow \ln {{y}_{1}}=\ln \left( {{\pi }^{x}} \right)$
We know that $\ln \left( {{a}^{b}} \right)$ is equal to $b\ln \left( a \right).$ Using this for the right-hand side,
$\Rightarrow \ln {{y}_{1}}=x.\ln \left( \pi \right)$
We know the differentiation of ln function can be given by $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}.$ Differentiating both sides of the above equation using this,
$\Rightarrow \dfrac{1}{{{y}_{1}}}.\dfrac{d{{y}_{1}}}{dx}=\dfrac{dx}{dx}.\ln \left( \pi \right)$
This is simplified to,
$\Rightarrow \dfrac{1}{{{y}_{1}}}.\dfrac{d{{y}_{1}}}{dx}=1.\ln \left( \pi \right)$
We multiply both sides of the above equation by ${{y}_{1}}$ .
$\Rightarrow \dfrac{d{{y}_{1}}}{dx}=\ln \left( \pi \right).{{y}_{1}}$
We know that ${{y}_{1}}={{\pi }^{x}}.$ Substituting this in the above equation,
$\Rightarrow \dfrac{d{{y}_{1}}}{dx}=\ln \left( \pi \right).{{\pi }^{x}}\ldots \ldots \left( 1 \right)$
We shall now differentiate the second equation ${{y}_{2}}={{x}^{\pi }}$ by using the power law given as $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}.$
$\Rightarrow \dfrac{d}{dx}\left( {{y}_{2}} \right)=\pi {{x}^{\pi -1}}\ldots \ldots \left( 2 \right)$
Adding the above two equations, we get the final result.
$\Rightarrow \dfrac{dy}{dx}=\ln \left( \pi \right).{{\pi }^{x}}+\pi {{x}^{\pi -1}}$
Hence, the derivative of $y={{\pi }^{x}}+{{x}^{\pi }}$ is $\ln \left( \pi \right).{{\pi }^{x}}+\pi {{x}^{\pi -1}}.$
Note: We need to know the basic differentiation formula in order to solve such questions. These form the basis for many mathematical problems. Students need to remember these formulas. We need to be careful while differentiating the ln function on the left-hand side for ${{y}_{1}}$ such that we get the $\dfrac{d{{y}_{1}}}{dx}$ term too. Students tend to miss out on this term and need to be careful.
Complete step by step answer:
In order to solve this question, let us split the given question $y={{\pi }^{x}}+{{x}^{\pi }}$ into two parts ${{y}_{1}}$ and ${{y}_{2}}.$ Let ${{y}_{1}}={{\pi }^{x}}$ and ${{y}_{2}}={{x}^{\pi }},$ and we need to differentiate the two equations separately. Let us consider the first equation,
$\Rightarrow {{y}_{1}}={{\pi }^{x}}$
Let us apply ln on both sides now,
$\Rightarrow \ln {{y}_{1}}=\ln \left( {{\pi }^{x}} \right)$
We know that $\ln \left( {{a}^{b}} \right)$ is equal to $b\ln \left( a \right).$ Using this for the right-hand side,
$\Rightarrow \ln {{y}_{1}}=x.\ln \left( \pi \right)$
We know the differentiation of ln function can be given by $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}.$ Differentiating both sides of the above equation using this,
$\Rightarrow \dfrac{1}{{{y}_{1}}}.\dfrac{d{{y}_{1}}}{dx}=\dfrac{dx}{dx}.\ln \left( \pi \right)$
This is simplified to,
$\Rightarrow \dfrac{1}{{{y}_{1}}}.\dfrac{d{{y}_{1}}}{dx}=1.\ln \left( \pi \right)$
We multiply both sides of the above equation by ${{y}_{1}}$ .
$\Rightarrow \dfrac{d{{y}_{1}}}{dx}=\ln \left( \pi \right).{{y}_{1}}$
We know that ${{y}_{1}}={{\pi }^{x}}.$ Substituting this in the above equation,
$\Rightarrow \dfrac{d{{y}_{1}}}{dx}=\ln \left( \pi \right).{{\pi }^{x}}\ldots \ldots \left( 1 \right)$
We shall now differentiate the second equation ${{y}_{2}}={{x}^{\pi }}$ by using the power law given as $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}.$
$\Rightarrow \dfrac{d}{dx}\left( {{y}_{2}} \right)=\pi {{x}^{\pi -1}}\ldots \ldots \left( 2 \right)$
Adding the above two equations, we get the final result.
$\Rightarrow \dfrac{dy}{dx}=\ln \left( \pi \right).{{\pi }^{x}}+\pi {{x}^{\pi -1}}$
Hence, the derivative of $y={{\pi }^{x}}+{{x}^{\pi }}$ is $\ln \left( \pi \right).{{\pi }^{x}}+\pi {{x}^{\pi -1}}.$
Note: We need to know the basic differentiation formula in order to solve such questions. These form the basis for many mathematical problems. Students need to remember these formulas. We need to be careful while differentiating the ln function on the left-hand side for ${{y}_{1}}$ such that we get the $\dfrac{d{{y}_{1}}}{dx}$ term too. Students tend to miss out on this term and need to be careful.
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