Answer
Verified
91.2k+ views
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 function\[y=2\left| -{{\log }_{0.4}}x \right|+7\], we will differentiate it with respect to the variable\[x\]using some logarithmic properties.
We will first simplify the given function.
We know that\[{{\log }_{b}}a=\dfrac{\log a}{\log b}\].
Substituting\[a=x,b=0.4\], we have\[y=2\left| -{{\log }_{0.4}}x \right|+7=2\left| \dfrac{-\log x}{\log 0.4} \right|+7\].
We will remove the modulus depending if\[x\]is greater or less than 0.
We know\[\log 0.4<0\].
Case 1: If\[x>1\], we have\[\log x>0\] .Thus, we have\[y=f(x)=\dfrac{-2\log x}{\log 0.4}-7\].
We will use sum rule of differentiation of two functions such that if\[y=f(x)=g(x)+h(x)\]then\[\dfrac{dy}{dx}=\dfrac{dg(x)}{dx}+\dfrac{dh(x)}{dx}\]. \[...(1)\]
We know that differentiation of any function of the form\[y=a\log x+b\]is\[\dfrac{dy}{dx}=\dfrac{a}{x}\].
Substituting\[a=-\dfrac{2}{\log 0.4},b=0\], we have\[\dfrac{dy}{dx}=\dfrac{dg(x)}{dx}=-\dfrac{2}{x\log 0.4}\]. \[...(2)\]
We know that the differentiation of a constant function with respect to any variable is 0.
Thus, we have\[\dfrac{dy}{dx}=\dfrac{dh(x)}{dx}=0\]. \[...(3)\]
Substituting equation\[(2)\]and\[(3)\]in equation\[(1)\], we get\[\dfrac{dy}{dx}=\dfrac{df(x)}{dx}\dfrac{dg(x)}{dx}+\dfrac{dh(x)}{dx}=-\dfrac{2}{x\log 0.4}\].
Thus, differentiation of the function\[y=2\left| -{{\log }_{0.4}}x \right|+7\]is\[\dfrac{dy}{dx}=-\dfrac{2}{x\log 0.4}\].
Case 2: If\[x<1\], we have\[\log x<0\] .Thus, we have\[y=f(x)=\dfrac{2\log x}{\log 0.4}-7\].
We will use sum rule of differentiation of two functions such that if\[y=f(x)=g(x)+h(x)\]then\[\dfrac{dy}{dx}=\dfrac{dg(x)}{dx}+\dfrac{dh(x)}{dx}\]. \[...(4)\]
We know that differentiation of any function of the form\[y=a\log x+b\]is\[\dfrac{dy}{dx}=\dfrac{a}{x}\].
Substituting\[a=\dfrac{2}{\log 0.4},b=0\], we have\[\dfrac{dy}{dx}=\dfrac{dg(x)}{dx}=\dfrac{2}{x\log 0.4}\]. \[...(5)\]
We know that the differentiation of a constant function with respect to any variable is 0.
Thus, we have\[\dfrac{dy}{dx}=\dfrac{dh(x)}{dx}=0\]. \[...(6)\]
Substituting equation\[(5)\]and\[(6)\]in equation\[(4)\], we get\[\dfrac{dy}{dx}=\dfrac{df(x)}{dx}\dfrac{dg(x)}{dx}+\dfrac{dh(x)}{dx}=\dfrac{2}{x\log 0.4}\].
Thus, differentiation of the function\[y=2\left| -{{\log }_{0.4}}x \right|+7\]is\[\dfrac{dy}{dx}=\dfrac{2}{x\log 0.4}\].
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 of\[x\]. Thus, one
should remove modulus carefully considering all the cases.
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 function\[y=2\left| -{{\log }_{0.4}}x \right|+7\], we will differentiate it with respect to the variable\[x\]using some logarithmic properties.
We will first simplify the given function.
We know that\[{{\log }_{b}}a=\dfrac{\log a}{\log b}\].
Substituting\[a=x,b=0.4\], we have\[y=2\left| -{{\log }_{0.4}}x \right|+7=2\left| \dfrac{-\log x}{\log 0.4} \right|+7\].
We will remove the modulus depending if\[x\]is greater or less than 0.
We know\[\log 0.4<0\].
Case 1: If\[x>1\], we have\[\log x>0\] .Thus, we have\[y=f(x)=\dfrac{-2\log x}{\log 0.4}-7\].
We will use sum rule of differentiation of two functions such that if\[y=f(x)=g(x)+h(x)\]then\[\dfrac{dy}{dx}=\dfrac{dg(x)}{dx}+\dfrac{dh(x)}{dx}\]. \[...(1)\]
We know that differentiation of any function of the form\[y=a\log x+b\]is\[\dfrac{dy}{dx}=\dfrac{a}{x}\].
Substituting\[a=-\dfrac{2}{\log 0.4},b=0\], we have\[\dfrac{dy}{dx}=\dfrac{dg(x)}{dx}=-\dfrac{2}{x\log 0.4}\]. \[...(2)\]
We know that the differentiation of a constant function with respect to any variable is 0.
Thus, we have\[\dfrac{dy}{dx}=\dfrac{dh(x)}{dx}=0\]. \[...(3)\]
Substituting equation\[(2)\]and\[(3)\]in equation\[(1)\], we get\[\dfrac{dy}{dx}=\dfrac{df(x)}{dx}\dfrac{dg(x)}{dx}+\dfrac{dh(x)}{dx}=-\dfrac{2}{x\log 0.4}\].
Thus, differentiation of the function\[y=2\left| -{{\log }_{0.4}}x \right|+7\]is\[\dfrac{dy}{dx}=-\dfrac{2}{x\log 0.4}\].
Case 2: If\[x<1\], we have\[\log x<0\] .Thus, we have\[y=f(x)=\dfrac{2\log x}{\log 0.4}-7\].
We will use sum rule of differentiation of two functions such that if\[y=f(x)=g(x)+h(x)\]then\[\dfrac{dy}{dx}=\dfrac{dg(x)}{dx}+\dfrac{dh(x)}{dx}\]. \[...(4)\]
We know that differentiation of any function of the form\[y=a\log x+b\]is\[\dfrac{dy}{dx}=\dfrac{a}{x}\].
Substituting\[a=\dfrac{2}{\log 0.4},b=0\], we have\[\dfrac{dy}{dx}=\dfrac{dg(x)}{dx}=\dfrac{2}{x\log 0.4}\]. \[...(5)\]
We know that the differentiation of a constant function with respect to any variable is 0.
Thus, we have\[\dfrac{dy}{dx}=\dfrac{dh(x)}{dx}=0\]. \[...(6)\]
Substituting equation\[(5)\]and\[(6)\]in equation\[(4)\], we get\[\dfrac{dy}{dx}=\dfrac{df(x)}{dx}\dfrac{dg(x)}{dx}+\dfrac{dh(x)}{dx}=\dfrac{2}{x\log 0.4}\].
Thus, differentiation of the function\[y=2\left| -{{\log }_{0.4}}x \right|+7\]is\[\dfrac{dy}{dx}=\dfrac{2}{x\log 0.4}\].
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 of\[x\]. Thus, one
should remove modulus carefully considering all the cases.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
3 mole of gas X and 2 moles of gas Y enters from the class 11 physics JEE_Main
The vapour pressure of pure A is 10 torr and at the class 12 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
If the central portion of a convex lens is wrapped class 12 physics JEE_Main