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Question: Find the derivative of the following function. \[y=2\left| -{{\log }_{0.4}}x \right|+7\]...

Find the derivative of the following function. y=2log0.4x+7y=2\left| -{{\log }_{0.4}}x \right|+7

Explanation

Solution

Hint: To find the derivative of the function given in the question, one must start by
simplifying the given function using properties of logarithmic function and then differentiating the terms given in the function using sum and product rule of differentiation.

To find the derivative of the functiony=2log0.4x+7y=2\left| -{{\log }_{0.4}}x \right|+7, we will differentiate it with respect to the variablexxusing some logarithmic properties.
We will first simplify the given function.
We know thatlogba=logalogb{{\log }_{b}}a=\dfrac{\log a}{\log b}.
Substitutinga=x,b=0.4a=x,b=0.4, we havey=2log0.4x+7=2logxlog0.4+7y=2\left| -{{\log }_{0.4}}x \right|+7=2\left| \dfrac{-\log x}{\log 0.4} \right|+7.
We will remove the modulus depending ifxxis greater or less than 0.
We knowlog0.4<0\log 0.4<0.
Case 1: Ifx>1x>1, we havelogx>0\log x>0 .Thus, we havey=f(x)=2logxlog0.47y=f(x)=\dfrac{-2\log x}{\log 0.4}-7.
We will use sum rule of differentiation of two functions such that ify=f(x)=g(x)+h(x)y=f(x)=g(x)+h(x)thendydx=dg(x)dx+dh(x)dx\dfrac{dy}{dx}=\dfrac{dg(x)}{dx}+\dfrac{dh(x)}{dx}. ...(1)...(1)
We know that differentiation of any function of the formy=alogx+by=a\log x+bisdydx=ax\dfrac{dy}{dx}=\dfrac{a}{x}.
Substitutinga=2log0.4,b=0a=-\dfrac{2}{\log 0.4},b=0, we havedydx=dg(x)dx=2xlog0.4\dfrac{dy}{dx}=\dfrac{dg(x)}{dx}=-\dfrac{2}{x\log 0.4}. ...(2)...(2)
We know that the differentiation of a constant function with respect to any variable is 0.
Thus, we havedydx=dh(x)dx=0\dfrac{dy}{dx}=\dfrac{dh(x)}{dx}=0. ...(3)...(3)
Substituting equation(2)(2)and(3)(3)in equation(1)(1), we getdydx=df(x)dxdg(x)dx+dh(x)dx=2xlog0.4\dfrac{dy}{dx}=\dfrac{df(x)}{dx}\dfrac{dg(x)}{dx}+\dfrac{dh(x)}{dx}=-\dfrac{2}{x\log 0.4}.
Thus, differentiation of the functiony=2log0.4x+7y=2\left| -{{\log }_{0.4}}x \right|+7isdydx=2xlog0.4\dfrac{dy}{dx}=-\dfrac{2}{x\log 0.4}.
Case 2: Ifx<1x<1, we havelogx<0\log x<0 .Thus, we havey=f(x)=2logxlog0.47y=f(x)=\dfrac{2\log x}{\log 0.4}-7.

We will use sum rule of differentiation of two functions such that ify=f(x)=g(x)+h(x)y=f(x)=g(x)+h(x)thendydx=dg(x)dx+dh(x)dx\dfrac{dy}{dx}=\dfrac{dg(x)}{dx}+\dfrac{dh(x)}{dx}. ...(4)...(4)

We know that differentiation of any function of the formy=alogx+by=a\log x+bisdydx=ax\dfrac{dy}{dx}=\dfrac{a}{x}.
Substitutinga=2log0.4,b=0a=\dfrac{2}{\log 0.4},b=0, we havedydx=dg(x)dx=2xlog0.4\dfrac{dy}{dx}=\dfrac{dg(x)}{dx}=\dfrac{2}{x\log 0.4}. ...(5)...(5)
We know that the differentiation of a constant function with respect to any variable is 0.
Thus, we havedydx=dh(x)dx=0\dfrac{dy}{dx}=\dfrac{dh(x)}{dx}=0. ...(6)...(6)
Substituting equation(5)(5)and(6)(6)in equation(4)(4), we getdydx=df(x)dxdg(x)dx+dh(x)dx=2xlog0.4\dfrac{dy}{dx}=\dfrac{df(x)}{dx}\dfrac{dg(x)}{dx}+\dfrac{dh(x)}{dx}=\dfrac{2}{x\log 0.4}.
Thus, differentiation of the functiony=2log0.4x+7y=2\left| -{{\log }_{0.4}}x \right|+7isdydx=2xlog0.4\dfrac{dy}{dx}=\dfrac{2}{x\log 0.4}.
Note: The first derivative of any function signifies the slope of the function. Also, we get
different values of derivatives of the function based on different values ofxx. Thus, one
should remove modulus carefully considering all the cases.