Compute the derivative of f(x)=5^x^2+1 using logarithmic differentiation.
Check using the following formula:
(5^g(x))'= (ln5)*(5^g(x))*(g'(x))
I never did like formulas.
remember that derivative of ln(g(x))=g'(x)/g(x)
here f(x)=5^(x^2+1)
ln f(x)= (x^2+1)ln5
take the derivative
f'(x)/f(x)=2x*ln5 + 0
f'(x)=f(x)*10x*ln5=10x*ln5*5^(x^2+1)
Now comparing this to your 'formula':
f'(x)=ln5*5^(x^2+1)*2x=10x*5^(x^2+1)
Hmmm same thing.
To compute the derivative of \(f(x) = 5^{x^{2} + 1}\) using logarithmic differentiation, follow these steps:
Step 1: Take the natural logarithm (ln) of both sides of the equation:
\[\ln(f(x)) = \ln(5^{x^{2}+1})\]
Step 2: Use the logarithm rule \(\ln(a^b) = b \cdot \ln(a)\) to simplify the right side of the equation:
\[\ln(f(x)) = (x^{2}+1) \cdot \ln(5)\]
Step 3: Differentiate both sides of the equation implicitly with respect to \(x\). The derivative of \(\ln(f(x))\) can be found using the chain rule:
\[\frac{1}{f(x)} \cdot f'(x) = 2x \cdot \ln(5)\]
Step 4: Solve the equation for \(f'(x)\):
\[f'(x) = f(x) \cdot 2x \cdot \ln(5)\]
Step 5: Substitute \(f(x) = 5^{x^{2}+1}\) back into the equation:
\[f'(x) = 5^{x^{2}+1} \cdot 2x \cdot \ln(5)\]
Therefore, the derivative of \(f(x) = 5^{x^{2} + 1}\) using logarithmic differentiation is \(f'(x) = 5^{x^{2}+1} \cdot 2x \cdot \ln(5)\).