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Question: Consider the following function, \[y={{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}}\] then \[\left...

Consider the following function, y=(1−x)αeαxy={{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}} then (1−x)y1=\left( 1-x \right){{y}_{1}}=
(a)αy\left( a \right)\alpha y
(b)αxy\left( b \right)\alpha xy
(c)xyy\left( c \right)\dfrac{xy}{y}
(d)−αxy\left( d \right)-\alpha xy

Explanation

Solution

Hint : We have y=(1−x)αeαxy={{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}} which is the product of two functions (1−x)α{{\left( 1-x \right)}^{\alpha }} and eαx.{{e}^{\alpha x}}. In order to find y1{{y}_{1}} we have to first find the first derivative of y with respect to x. We use the product formula, i.e. d(uv)dx=vdudx+udvdx\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=(1−x)αeαx.y={{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}}. We have to calculate the value of (1−x)y1.\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 y1=dydx.{{y}_{1}}=\dfrac{dy}{dx}.
Now, as we can see that y=(1−x)αeαxy={{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}} is a product of two functions, so to find the derivative, we use the product rule,
d(uv)dx=vdudx+udvdx\dfrac{d\left( uv \right)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}
So we use this on y=(1−x)αeαx.y={{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}}.
So, we have,
dydx=d((1−x)αeαx)dx\dfrac{dy}{dx}=\dfrac{d\left( {{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}} \right)}{dx}
Using the product rule, we get,
⇒dydx=eαxddx(1−x)α+(1−x)αd(eαx)dx......(i)\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 d(xn)dx=nxn−1\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} and d(ex)dx=1,\dfrac{d\left( {{e}^{x}} \right)}{dx}=1, we get,
d(1−x)αdx=a(1−x)(−1)\dfrac{d{{\left( 1-x \right)}^{\alpha }}}{dx}=a\left( 1-x \right)\left( -1 \right)
And,
d(eαx)dx=eαxd(αx)αx\dfrac{d\left( {{e}^{\alpha x}} \right)}{dx}=\dfrac{{{e}^{\alpha x}}d\left( \alpha x \right)}{\alpha x}
⇒d(eαx)dx=eαx.α\Rightarrow \dfrac{d\left( {{e}^{\alpha x}} \right)}{dx}={{e}^{\alpha x}}.\alpha
⇒d(eαx)dx=α.eαx\Rightarrow \dfrac{d\left( {{e}^{\alpha x}} \right)}{dx}=\alpha .{{e}^{\alpha x}}
Now we put these two values in (i), we will have,
dydx=eαx[α(1−x)α−1(−1)]+(1−x)a(αeαx)\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,
dydx=−α(1−x)a−1eαx+α(1−x)αeαx\dfrac{dy}{dx}=-\alpha {{\left( 1-x \right)}^{a-1}}{{e}^{\alpha x}}+\alpha {{\left( 1-x \right)}^{\alpha }}{{e}^{\alpha x}}
Taking αeαx\alpha {{e}^{\alpha x}} we get,
⇒dydx=αeαx[−(1−x)a−1+(1−x)α]\Rightarrow \dfrac{dy}{dx}=\alpha {{e}^{\alpha x}}\left[ -{{\left( 1-x \right)}^{a-1}}+{{\left( 1-x \right)}^{\alpha }} \right]
Taking (1−x)α−1{{\left( 1-x \right)}^{\alpha -1}} common, we will get,
⇒dydx=αeαx(1−x)α−1[−1+1−x]\Rightarrow \dfrac{dy}{dx}=\alpha {{e}^{\alpha x}}{{\left( 1-x \right)}^{\alpha -1}}\left[ -1+1-x \right]
On simplifying, we will get,
⇒dydx=αeαx(1−x)α−1(−x)\Rightarrow \dfrac{dy}{dx}=\alpha {{e}^{\alpha x}}{{\left( 1-x \right)}^{\alpha -1}}\left( -x \right)
So, we get,
⇒y1=dydx=αeαx(1−x)α−1(−x).......(ii)\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 (1−x)y1.\left( 1-x \right){{y}_{1}}. So, we will multiply the value of y1{{y}_{1}} in (ii) by (1 – x). So, we get,
⇒(1−x)y1=αeαx(1−x)α−1(−x)(1−x)\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, eαx(1−x)α=y,{{e}^{\alpha x}}{{\left( 1-x \right)}^{\alpha }}=y, we get,
⇒(1−x)y1=αeαx(1−x)α(−x)\Rightarrow \left( 1-x \right){{y}_{1}}=\alpha {{e}^{\alpha x}}{{\left( 1-x \right)}^{\alpha }}\left( -x \right)
⇒(1−x)y1=αy(−x)\Rightarrow \left( 1-x \right){{y}_{1}}=\alpha y\left( -x \right)
⇒(1−x)y1=−xαy\Rightarrow \left( 1-x \right){{y}_{1}}=-x\alpha y
So we get the required answer as −xαy.-x\alpha y.
So, the correct answer is “Option D”.

Note : While finding the derivative, always double-check your solution, as d[(1−x)α]dx≠α(1−x)α−1.\dfrac{d\left[ {{\left( 1-x \right)}^{\alpha }} \right]}{dx}\ne \alpha {{\left( 1-x \right)}^{\alpha -1}}. When we differentiate (1−x)α{{\left( 1-x \right)}^{\alpha }} with x, first we take (1 – x) as t. So, d(1−x)αdx=d(tα)dx=αtα−1dtdx.\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.
⇒d(t)dx=d(1−x)dx\Rightarrow \dfrac{d\left( t \right)}{dx}=\dfrac{d\left( 1-x \right)}{dx}
⇒d(t)dx=d(1)dx−d(x)dx\Rightarrow \dfrac{d\left( t \right)}{dx}=\dfrac{d\left( 1 \right)}{dx}-\dfrac{d\left( x \right)}{dx}
⇒d(t)dx=−1\Rightarrow \dfrac{d\left( t \right)}{dx}=-1
So, we get,
d[(1−x)α]dx=α(1−x)α−1(−1)\dfrac{d\left[ {{\left( 1-x \right)}^{\alpha }} \right]}{dx}=\alpha {{\left( 1-x \right)}^{\alpha -1}}\left( -1 \right)