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$

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.

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,
$\dfrac{d\left[ {{\left( 1-x \right)}^{\alpha }} \right]}{dx}=\alpha {{\left( 1-x \right)}^{\alpha -1}}\left( -1 \right)$