Question

How do you find the derivative of \[y=20{{x}^{\dfrac{1}{4}}}-3{{x}^{\dfrac{3}{5}}}-2{{e}^{x}}\]?

Answer 1

Hint: In mathematics the derivative of function defined as if we have $x$ and $y$ real numbers, and if the graph of f is plotted with respect to $x$ then the derivative of the function is the slope of this graph at each points. Here the $x$ is independent variable and $y$ is the dependent variable. These types of questions are based on the concept of differentiation. To find the derivative of such types of questions in the above equation \[y=20{{x}^{\dfrac{1}{4}}}-3{{x}^{\dfrac{3}{5}}}-2{{e}^{x}}\] we will use the power rule of differentiation. The power rule says if we have any polynomial function like ${{x}^{n}}$ then the differentiation of function is as \[\dfrac{d\left[ a{{x}^{n}} \right]}{dy}=na{{x}^{n-1}}\] . Where $a$ is the coefficient of $x$ and $n$ is the power of the $x$.

Complete step by step solution:
Here the given equation is:
\[y=20{{x}^{\dfrac{1}{4}}}-3{{x}^{\dfrac{3}{5}}}-2{{e}^{x}}\]
The value of $y$ changes as $x$ changes.
Now we will use power rule\ [\dfrac {d\left [a {{x} ^ {n}} \right]}{dy}=na{{x}^{n-1}}\] on given functions one by one:
 $\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( 20x\dfrac{1}{4} \right)}{dx}-\dfrac{d\left( 3{{x}^{\dfrac{3}{5}}} \right)}{dx}-\dfrac{d\left( 2{{e}^{x}} \right)}{dx}.......\left( 1 \right)$
Now by comparing the $\dfrac{d\left( 20{{x}^{\dfrac{1}{4}}} \right)}{dx}$ with the LHS side of the formula of the power rule \[\dfrac{d\left[ a{{x}^{n}} \right]}{dy}=na{{x}^{n-1}}\]then we get,$a=20$ and $n=\dfrac{1}{4}$
Now by comparing with the RHS the differentiation of $\dfrac{d\left( 20{{x}^{\dfrac{1}{4}}} \right)}{dx}$ $=\dfrac{20}{4}{{x}^{\dfrac{1}{4}-1}}=5{{x}^{\dfrac{1-4}{4}}}=5{{x}^{\dfrac{-3}{4}}}........\left( 2 \right)$
Similarly by comparing the$\dfrac{d\left( 3{{x}^{\dfrac{3}{5}}} \right)}{dx}$ with the LHS of the power formula\[\dfrac{d\left[ a{{x}^{n}} \right]}{dy}=na{{x}^{n-1}}\]then we get, $a=3$ and $x=\dfrac{3}{5}$
By comparing with RHS the differentiation of $\dfrac{d\left( 3{{x}^{\dfrac{3}{5}}} \right)}{dx}=3\left( \dfrac{3}{5} \right){{x}^{\dfrac{3}{5}-1}}=\dfrac{9}{5}{{x}^{\dfrac{3-5}{5}}}=\dfrac{9}{5}{{x}^{\dfrac{-2}{5}}}.......\left( 3 \right)$
As we all know the differentiation of the ${{e} ^ {1\cdot x}} $ is equal to$1\cdot {{e} ^{x}} $
And differentiation of $\dfrac {d\left (2{{e} ^{x}} \right)}{dx}=2{{e}^{x}}.......\left (4 \right)$
Now putting all the values from (2), (3) and (4) in equation (1), we get:
$\Rightarrow \dfrac{dy}{dx}=5{{x}^{\dfrac{-3}{4}}}-\dfrac{9}{5}{{x}^{\dfrac{-2}{5}}}-2{{e}^{x}}$
Hence simply by using the power rule of differentiation\[\dfrac{d\left[ a{{x}^{n}} \right]}{dy}=na{{x}^{n-1}}\] we get the derivative of the \[y=20{{x}^{\dfrac{1}{4}}}-3{{x}^{\dfrac{3}{5}}}-2{{e}^{x}}\] is \[5{{x}^{\dfrac{-3}{4}}}-9{{x}^{\dfrac{-2}{5}}}-2{{e}^{x}}\].

Note:
The derivative of any equation is easy to find. Sometimes we can make mistakes by taking wrong differentiation of$n {{x} ^ {n-1}} $. In the place of (n-1) we make mistakes by writing (n+1) so it's mandatory for us to know the differentiation of every function used in an equation.