Question

Consider the following function, \[y={{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}}\] then \[\left( 1-x \right){{y}_{1}}=\]
\[\left( a \right)\alpha y\]
\[\left( b \right)\alpha xy\]
\[\left( c \right)\dfrac{xy}{y}\]
\[\left( d \right)-\alpha xy\]

Answer 1

Hint: We have \[y={{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}}\] which is the product of two functions \[{{\left( 1-x \right)}^{\alpha }}\] and \[{{e}^{\alpha x}}.\] In order to find \[{{y}_{1}}\] we have to first find the first derivative of y with respect to x. We use the product formula, i.e. \[\dfrac{d\left( uv \right)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\] to find the first derivative. When we find our first derivative, we multiply it with (1 – x) to get our required solution.

Complete step-by-step answer:
We are given y as \[y={{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}}.\] We have to calculate the value of \[\left( 1-x \right){{y}_{1}}.\] To so, we have to evaluate the first derivative of y with respect to x as we know that \[{{y}_{1}}=\dfrac{dy}{dx}.\]
Now, as we can see that \[y={{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}}\] is a product of two functions, so to find the derivative, we use the product rule,
\[\dfrac{d\left( uv \right)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\]
So we use this on \[y={{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}}.\]
So, we have,
\[\dfrac{dy}{dx}=\dfrac{d\left( {{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}} \right)}{dx}\]
Using the product rule, we get,
\[\Rightarrow \dfrac{dy}{dx}={{e}^{\alpha x}}\dfrac{d}{dx}{{\left( 1-x \right)}^{\alpha }}+{{\left( 1-x \right)}^{\alpha }}\dfrac{d\left( {{e}^{\alpha x}} \right)}{dx}......\left( i \right)\]
Now as \[\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}\] and \[\dfrac{d\left( {{e}^{x}} \right)}{dx}=1,\] we get,
\[\dfrac{d{{\left( 1-x \right)}^{\alpha }}}{dx}=a\left( 1-x \right)\left( -1 \right)\]
And,
\[\dfrac{d\left( {{e}^{\alpha x}} \right)}{dx}=\dfrac{{{e}^{\alpha x}}d\left( \alpha x \right)}{\alpha x}\]
\[\Rightarrow \dfrac{d\left( {{e}^{\alpha x}} \right)}{dx}={{e}^{\alpha x}}.\alpha \]
\[\Rightarrow \dfrac{d\left( {{e}^{\alpha x}} \right)}{dx}=\alpha .{{e}^{\alpha x}}\]
Now we put these two values in (i), we will have,
\[\dfrac{dy}{dx}={{e}^{\alpha x}}\left[ \alpha {{\left( 1-x \right)}^{\alpha -1}}\left( -1 \right) \right]+{{\left( 1-x \right)}^{^{a}}}\left( \alpha {{e}^{\alpha x}} \right)\]
Now, simplifying we get,
\[\dfrac{dy}{dx}=-\alpha {{\left( 1-x \right)}^{a-1}}{{e}^{\alpha x}}+\alpha {{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}}\]
Taking \[\alpha {{e}^{\alpha x}}\] we get,
\[\Rightarrow \dfrac{dy}{dx}=\alpha {{e}^{\alpha x}}\left[ -{{\left( 1-x \right)}^{a-1}}+{{\left( 1-x \right)}^{\alpha }} \right]\]
 Taking \[{{\left( 1-x \right)}^{\alpha -1}}\] common, we will get,
\[\Rightarrow \dfrac{dy}{dx}=\alpha {{e}^{\alpha x}}{{\left( 1-x \right)}^{\alpha -1}}\left[ -1+1-x \right]\]
On simplifying, we will get,
\[\Rightarrow \dfrac{dy}{dx}=\alpha {{e}^{\alpha x}}{{\left( 1-x \right)}^{\alpha -1}}\left( -x \right)\]
So, we get,
\[\Rightarrow {{y}_{1}}=\dfrac{dy}{dx}=\alpha {{e}^{\alpha x}}{{\left( 1-x \right)}^{\alpha -1}}\left( -x \right).......\left( ii \right)\]
Now, we want to find the value \[\left( 1-x \right){{y}_{1}}.\] So, we will multiply the value of \[{{y}_{1}}\] in (ii) by (1 – x). So, we get,
\[\Rightarrow \left( 1-x \right){{y}_{1}}=\alpha {{e}^{\alpha x}}{{\left( 1-x \right)}^{\alpha -1}}\left( -x \right)\left( 1-x \right)\]
As, \[{{e}^{\alpha x}}{{\left( 1-x \right)}^{\alpha }}=y,\] we get,
\[\Rightarrow \left( 1-x \right){{y}_{1}}=\alpha {{e}^{\alpha x}}{{\left( 1-x \right)}^{\alpha }}\left( -x \right)\]
\[\Rightarrow \left( 1-x \right){{y}_{1}}=\alpha y\left( -x \right)\]
\[\Rightarrow \left( 1-x \right){{y}_{1}}=-x\alpha y\]
So we get the required answer as \[-x\alpha y.\]
So, the correct answer is “Option D”.

Note: While finding the derivative, always double-check your solution, as \[\dfrac{d\left[ {{\left( 1-x \right)}^{\alpha }} \right]}{dx}\ne \alpha {{\left( 1-x \right)}^{\alpha -1}}.\] When we differentiate \[{{\left( 1-x \right)}^{\alpha }}\] with x, first we take (1 – x) as t. So, \[\dfrac{d{{\left( 1-x \right)}^{\alpha }}}{dx}=\dfrac{d\left( {{t}^{\alpha }} \right)}{dx}=\alpha {{t}^{\alpha -1}}\dfrac{dt}{dx}.\] Now, we will differentiate t = 1 – x with respect to x.
\[\Rightarrow \dfrac{d\left( t \right)}{dx}=\dfrac{d\left( 1-x \right)}{dx}\]
\[\Rightarrow \dfrac{d\left( t \right)}{dx}=\dfrac{d\left( 1 \right)}{dx}-\dfrac{d\left( x \right)}{dx}\]
\[\Rightarrow \dfrac{d\left( t \right)}{dx}=-1\]
So, we get,
\[\dfrac{d\left[ {{\left( 1-x \right)}^{\alpha }} \right]}{dx}=\alpha {{\left( 1-x \right)}^{\alpha -1}}\left( -1 \right)\]