Question

d(2 ka power x)/d (3 ka power x)=​

Answer 1

Question:

Differentiate 2ˣ with respect to 3ˣ i. e. find

$\displaystyle{\sf\:\dfrac{d\:(\:2^x\:)}{d\:(\:3^x\:)}}$

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{d\:(\:2^x\:)}{d\:(\:3^x\:)}\:=\:\dfrac{\ln\:(\:2\:)\:2^x}{\ln\:(\:3\:)\:3^x}\:}}}$

Step-by-step-explanation:

We have given two functions 2ˣ and 3ˣ.

We have to differentiate the first function with respect to the second.

Now, differentiating 2ˣ with respect to x,

$\displaystyle{\sf\:\dfrac{d\:(\:2^x\:)}{dx}}$

$\displaystyle{\implies\sf\:\dfrac{d\:(\:e^{\ln\:(\:2\:)\:x}\:)}{dx}}$

$\displaystyle{\implies\sf\:e^{\ln\:(\:2\:)\:x}\:.\:\dfrac{d\:(\:\ln\:(\:2\:)\:x\:)}{dx}}$

$\displaystyle{\implies\sf\:e^{\ln\:(\:2\:)\:x}\:.\:\ln\:(\:2\:)\:\dfrac{d\:(\:x\:)}{dx}}$

$\displaystyle{\implies\sf\:e^{\ln\:(\:2\:)\:x}\:.\:\ln\:(\:2\:)\:\times\:1}$

$\displaystyle{\implies\sf\:2^x\:\ln\:(\:2\:)}$

$\displaystyle{\therefore\:\boxed{\green{\sf\:\dfrac{d\:(\:2^x\:)}{dx}\:=\:\ln\:(\:2\:)\:2^x\:}}}$

Similarly, we can find the derivative of 3ˣ with respect to x.

$\displaystyle{\boxed{\blue{\sf\:\dfrac{d\:(\:3^x\:)}{dx}\:=\:\ln\:(\:3\:)\:3^x}}}$

Now, we know that,

$\displaystyle{\pink{\sf\:\dfrac{dy}{du}\:=\:\dfrac{dy}{dx}\:\times\:\dfrac{dx}{du}}}$

$\displaystyle{\implies\sf\:\dfrac{d\:(\:2^x\:)}{d\:(\:3^x\:)}\:=\:\dfrac{d\:(\:2^x\:)}{dx}\:\times\:\dfrac{dx}{d\:(\:3^x\:)}}$

$\displaystyle{\implies\sf\:\dfrac{d\:(\:2^x\:)}{d\:(\:3^x\:)}\:=\:\dfrac{d\:(\:2^x\:)}{dx}\:\times\:\dfrac{1}{\dfrac{d\:(\:3^x\:)}{dx}}}$

$\displaystyle{\implies\sf\:\dfrac{d\:(\:2^x\:)}{d\:(\:3^x\:)}\:=\:\ln\:(\:2\:)\:2^x\:\times\:\dfrac{1}{\ln\:(\:3\:)\:3^x}}$

$\displaystyle{\therefore\underline{\boxed{\red{\sf\:\dfrac{d\:(\:2^x\:)}{d\:(\:3^x\:)}\:=\:\dfrac{\ln\:(\:2\:)\:2^x}{\ln\:(\:3\:)\:3^x}\:}}}}$

Answer 2

QuEsTiOn,

• d(2 ka power x)/d (3 ka power x)=

To FiNd,

$\color{red}\displaystyle{\sf\:\dfrac{d\:(\:2^x\:)}{d\:(\:3^x\:)}}$

SoLuTiOn,

$\displaystyle{\boxed{\pink{\sf\:\dfrac{d\:(\:2^x\:)}{d\:(\:3^x\:)}\:=\:\dfrac{\ln\:(\:2\:)\:2^x}{\ln\:(\:3\:)\:3^x}\:}}} </p><p>$

We have given two functions 2ˣ and 3ˣ.

We have to differentiate the first function with respect to the second.

Now, differentiating 2ˣ with respect to x,

$\displaystyle{\sf\:\dfrac{d\:(\:2^x\:)}{dx}} </p><p></p><p> </p><p></p><p> \\ \\ \displaystyle{\implies\sf\:\dfrac{d\:(\:e^{\ln\:(\:2\:)\:x}\:)}{dx}}</p><p></p><p> </p><p></p><p> \\ \\ \displaystyle{\implies\sf\:e^{\ln\:(\:2\:)\:x}\:.\:\dfrac{d\:(\:\ln\:(\:2\:)\:x\:)}{dx}}</p><p></p><p> </p><p></p><p> \\ \\ \displaystyle{\implies\sf\:e^{\ln\:(\:2\:)\:x}\:.\:\ln\:(\:2\:)\:\dfrac{d\:(\:x\:)}{dx}}$

$\displaystyle{\implies\sf\:e^{\ln\:(\:2\:)\:x}\:.\:\ln\:(\:2\:)\:\times\:1}</p><p> \\ \\ \displaystyle{\implies\sf\:2^x\:\ln\:(\:2\:)}</p><p> \\ \\ \displaystyle{\therefore\:\boxed{\pink{\sf\:\dfrac{d\:(\:2^x\:)}{dx}\:=\:\ln\:(\:2\:)\:2^x\:}}}$

Similarly, we can find the derivative of 3ˣ with respect to x.

$\displaystyle{\boxed{\pink{\sf\:\dfrac{d\:(\:3^x\:)}{dx}\:=\:\ln\:(\:3\:)\:3^x}}} </p><p>$

Now, we know that,

$\displaystyle{\blue{\sf\:\dfrac{dy}{du}\:=\:\dfrac{dy}{dx}\:\times\:\dfrac{dx}{du}}} </p><p></p><p> </p><p></p><p> \\ \\ \displaystyle{\implies\sf\:\dfrac{d\:(\:2^x\:)}{d\:(\:3^x\:)}\:=\:\dfrac{d\:(\:2^x\:)}{dx}\:\times\:\dfrac{dx}{d\:(\:3^x\:)}}$

$\displaystyle{\implies\sf\:\dfrac{d\:(\:2^x\:)}{d\:(\:3^x\:)}\:=\:\dfrac{d\:(\:2^x\:)}{dx}\:\times\:\dfrac{1}{\dfrac{d\:(\:3^x\:)}{dx}}}</p><p></p><p> \\ \\ \displaystyle{\implies\sf\:\dfrac{d\:(\:2^x\:)}{d\:(\:3^x\:)}\:=\:\ln\:(\:2\:)\:2^x\:\times\:\dfrac{1}{\ln\:(\:3\:)\:3^x}}$

FiNaL AnSwEr,

$\displaystyle{\therefore\underline{\boxed{\pink{\sf\:\dfrac{d\:(\:2^x\:)}{d\:(\:3^x\:)}\:=\:\dfrac{\ln\:(\:2\:)\:2^x}{\ln\:(\:3\:)\:3^x}\:}}}}$

Hope you get your AnSwEr.