If e^(x+3logx) then prove that dy/dx=x^2(x+3)e^x....can anyone help me out...plz note 3+logx is in adition with x and is the power of 'e' its not e^x+3logx
that is true, if 'ln' is used, not 'log'. Well, in some books, ln and log seem interchangeable -they really mean ln, not log.
If we rewrite the expression using ln, not log,
e^(x + 3lnx)
we can separate this into:
(e^x)[e^(3lnx)]
note that we can further rewrite and simplify the e^(3lnx) into e^(ln x^3) = x^3. Thus, we have
(e^x)(x^3)
and we use chain rule:
d/dx (f(x)g(x)) = f'(x)g(x) + f(x)g'(x)
which means, if you have two factors which are functions of x (in the problem, they are e^x and x^3), we get the derivative of the first factor then multiply to the second factor, plus the derivative of the second factor then multiply to first factor. Note that
d/dx (e^x) = e^x
d/dx (x^3) = 3x^2
Thus, the derivative is
(e^x)(x^3) + (e^x)(3x^2)
(e^x)(x^2)(x + 3)
hope this helps~ :)
To prove that dy/dx = x^2(x+3)e^x, we need to differentiate the given function with respect to x. Let's break down the function and differentiate each term step by step.
Given function: y = e^(x + 3logx)
Step 1: Express the function in exponential form
y = e^x * e^(3logx)
Step 2: Simplify the expression
Since e^(logx) = x (by definition of logarithms and exponential functions), we can rewrite the function as follows:
y = e^x * (e^logx)^3
Step 3: Further simplify the expression
Using the property that e^a * e^b = e^(a+b), we can simplify the equation as:
y = e^x * x^3
Step 4: Differentiate both sides of the equation with respect to x
(dy/dx) = (d/dx) [e^x * x^3]
Step 5: Apply the product rule of differentiation
(dy/dx) = (d/dx) [e^x] * x^3 + e^x * (d/dx) [x^3]
Step 6: Differentiate each term separately
(dy/dx) = e^x * x^3 + e^x * 3x^2
Step 7: Simplify the expression
(dy/dx) = x^3e^x + 3x^2e^x
Step 8: Factor out the common term e^x
(dy/dx) = (x^3 + 3x^2) * e^x
Finally, we have proven that dy/dx = x^2(x + 3)e^x, as required.