derivative of lim x->infinite (2^x+5^x)^(1/x)?
I do it till [2^xln(2)+5^xln(5)]/(2^x+5^x) and then I can't simplify anymore. Help me out?
We'll first find the limit, and I'll let you continue with the derivative.
Let
Y=Lim x->∞ (2^x+5^x)^(1/x)
=Lim x->∞ ((5*2/5)^x+5^x)^(1/x)
=Lim x->∞ ((5^x)(2/5)^x+5^x)^(1/x) [law of exponents]
=Lim x->∞ ((5^x)((2/5)^x+1)^(1/x) [factorize]
Take (natural) log on both sides:
log(Y)=Lim x->∞ (1/x)(xlog(5)+log(1+(2/5)^x)
Take limit x->∞,
(2/5)^x -> 0, hence log(1)=0
so
log(Y)=log(5)
=> Y=5=Lim x->∞ (2^x+5^x)^(1/x)
I don't get what you did after log(Y)=Lim x->∞ (1/x)(xlog(5)+log(1+(2/5)^x) . Why is (2/5)^x -> 0, and hence log(1)=0??
To find the derivative of the given expression, we need to rewrite the expression in a form that can be easily differentiated.
Let's start by taking the natural logarithm (ln) of the entire expression:
ln(y) = ln[(2^x + 5^x)^(1/x)]
Now, using the logarithmic identities, we can simplify the expression further:
ln(y) = (1/x) * ln(2^x + 5^x)
Next, we can apply the logarithmic property that states log(a^b) = b * log(a):
ln(y) = (1/x) * [x * ln(2) + x * ln(5)]
Now, let's simplify this:
ln(y) = ln(2) + ln(5)
Finally, we can raise both sides of the equation to the power of e (to remove the natural logarithm):
y = e^[ln(2) + ln(5)] = e^ln(2) * e^ln(5) = 2 * 5 = 10
Therefore, the original expression evaluates to a constant value of 10.
Since the derivative of a constant is always zero, the derivative of lim (x->∞) (2^x + 5^x)^(1/x) is equal to 0.