
How do you find the derivative of $y=6^{2x}$ ?
Answer
535.8k+ views
Hint: We start solving the problem by applying the chain rule $\dfrac{dy}{dx}=\left( \dfrac{dy}{dz} \right)\left( \dfrac{dz}{dx} \right)$ in the given equation. Then you solve the derivatives separately. Then we should multiply both the derivatives to get the required answer.
Complete step by step solution:
According to the problem, we are asked to find the derivative of the equation $y=6^{2x}$--- ( 1 )
Therefore, by using the chain rule $\dfrac{dy}{dx}=\left( \dfrac{dy}{dz} \right)\left( \dfrac{dz}{dx} \right)$, in equation 1, we get:
Here, let z = 2x
First, we calculate for $\dfrac{dy}{dz}$.
Therefore, $\dfrac{dy}{dz}=\dfrac{d{{6}^{z}}}{dz}$ [Since z = 2x]
To find the derivative of ${{6}^{z}}$, we can do the following:
Take y = ${{6}^{z}}$--- (2),
Now, apply ln on both sides.
$\Rightarrow \ln{y} = \ln{{6}^{z}}$
$\Rightarrow \ln{y} =z\ln{6}$
Now, let us differentiate both sides implicitly with respect to z
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\ln{6}$
$\Rightarrow \dfrac{dy}{dx}=y\ln{6}$
From equation (2), we know that y = ${{6}^{z}}$
Therefore, putting it in the above equation, we get
$\dfrac{dy}{dz}={{6}^{z}}\ln 6$ ----(3)
Now, let us calculate for $\dfrac{dz}{dx}$.
Let us again put z = 2x, in the above equation, that is $\dfrac{dz}{dx}$. Therefore, we get
$\Rightarrow \dfrac{dz}{dx}=\dfrac{d\left(2x\right)}{dx}$
$\Rightarrow \dfrac{dz}{dx}=2$ ----(3)
Since, we calculated both $\dfrac{dz}{dx}$ and $\dfrac{dy}{dz}$, let us substitute both the equations (2) and (3) in equation (1), that is $\dfrac{dy}{dx}=\left( \dfrac{dy}{dz} \right)\left( \dfrac{dz}{dx} \right)$.
$\Rightarrow \dfrac{dy}{dx}=({{6}^{z}}(\ln 6))(2)$ ----(4)
But now, we should substitute, z = 2x. Therefore, let us substitute equation (4) with z = 2x.
$\Rightarrow \dfrac{dy}{dx}=({{6}^{2x}}(\ln 6))(2)$
$\Rightarrow \dfrac{dy}{dx}=2.{{6}^{2x}}(\ln 6)$
So, we have found the derivative of the given equation $y=6^{2x}$ as $\dfrac{dy}{dx}={{2.6}^{2x}}(\ln 6)$
Therefore, the solution of the given equation $y=6^{2x}$ is $\dfrac{dy}{dx}={{2.6}^{2x}}(\ln 6)$
Note:
If we get these types of problems, we have to be careful while doing the substitutions for the equations.Also, we could directly use the formula of $\dfrac{d({{a}^{x}})}{dx}={{a}^{x}}(\ln a)$ while calculating the second equation.
Complete step by step solution:
According to the problem, we are asked to find the derivative of the equation $y=6^{2x}$--- ( 1 )
Therefore, by using the chain rule $\dfrac{dy}{dx}=\left( \dfrac{dy}{dz} \right)\left( \dfrac{dz}{dx} \right)$, in equation 1, we get:
Here, let z = 2x
First, we calculate for $\dfrac{dy}{dz}$.
Therefore, $\dfrac{dy}{dz}=\dfrac{d{{6}^{z}}}{dz}$ [Since z = 2x]
To find the derivative of ${{6}^{z}}$, we can do the following:
Take y = ${{6}^{z}}$--- (2),
Now, apply ln on both sides.
$\Rightarrow \ln{y} = \ln{{6}^{z}}$
$\Rightarrow \ln{y} =z\ln{6}$
Now, let us differentiate both sides implicitly with respect to z
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\ln{6}$
$\Rightarrow \dfrac{dy}{dx}=y\ln{6}$
From equation (2), we know that y = ${{6}^{z}}$
Therefore, putting it in the above equation, we get
$\dfrac{dy}{dz}={{6}^{z}}\ln 6$ ----(3)
Now, let us calculate for $\dfrac{dz}{dx}$.
Let us again put z = 2x, in the above equation, that is $\dfrac{dz}{dx}$. Therefore, we get
$\Rightarrow \dfrac{dz}{dx}=\dfrac{d\left(2x\right)}{dx}$
$\Rightarrow \dfrac{dz}{dx}=2$ ----(3)
Since, we calculated both $\dfrac{dz}{dx}$ and $\dfrac{dy}{dz}$, let us substitute both the equations (2) and (3) in equation (1), that is $\dfrac{dy}{dx}=\left( \dfrac{dy}{dz} \right)\left( \dfrac{dz}{dx} \right)$.
$\Rightarrow \dfrac{dy}{dx}=({{6}^{z}}(\ln 6))(2)$ ----(4)
But now, we should substitute, z = 2x. Therefore, let us substitute equation (4) with z = 2x.
$\Rightarrow \dfrac{dy}{dx}=({{6}^{2x}}(\ln 6))(2)$
$\Rightarrow \dfrac{dy}{dx}=2.{{6}^{2x}}(\ln 6)$
So, we have found the derivative of the given equation $y=6^{2x}$ as $\dfrac{dy}{dx}={{2.6}^{2x}}(\ln 6)$
Therefore, the solution of the given equation $y=6^{2x}$ is $\dfrac{dy}{dx}={{2.6}^{2x}}(\ln 6)$
Note:
If we get these types of problems, we have to be careful while doing the substitutions for the equations.Also, we could directly use the formula of $\dfrac{d({{a}^{x}})}{dx}={{a}^{x}}(\ln a)$ while calculating the second equation.
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