Find the derivative of the given function.g(t)=log5(7t2+4)g\left(t\right)=\log_5\left(7^{t^2+4}\right)g(t)=log5(7t2+4)

2 t ln ⁡ 7 ln ⁡ 5 \frac{2t\ln7}{\ln5} l n 5 2 t l n 7

Find the derivative of the given function. g ( t ) = log 5 ( 7 t 2 + 4 ) g\left(t\right)=\log_5\left(7^{t^2+4}\right) g ( t ) = lo g 5 ( 7 t 2 + 4 )

In this problem, we wanna find the derivative of the function g of t equals log base 5 of 7 to the power of t squared plus 4. Now here, we can clearly see we're not just taking the log of a variable. We're taking the log of an exponential function that is then raised to another function. Because of that, we know that we're gonna have to apply the chain rule here. Most of you have already learned the chain rule, but if you have not yet learned the chain rule, make sure that you circle back to this problem after you've gone through that. For the rest of you, let's get into things. We We wanna find our derivative here, g prime of t. Now starting with our outermost function here as we'll want to do with our chain rule, we have log base 5. So our derivative here is gonna be 1 over whatever is in our argument here, which is 7 to the power of t squared plus 4 times the natural log of our base, in this case, 5. Then working our way inside, we have this exponential function, 7 to the power of t squared plus 4. Now the derivative of this exponential function using our rule for general exponential functions is gonna be 7 to the power of t squared plus 4 times the natural log of the base of our exponent here, which in this case is 7, so natural log of 7. Then working our way to our innermost function, that's that t squared plus 4. What's the derivative of t squared plus 4? It is 2 t, so we're multiplying here by 2 t. Now here, I can see I have a 7 to the power of t squared plus 4 up top and on the bottom. So these are going to nicely cancel each other out here, and this then leaves me on the top with 2 t times the natural log of 7 over the natural log of 5. So this is my full derivative here, g prime of t, having applied the chain rule. Let me know if you have any questions here, and let's keep practicing.

Find the derivative of the given function. g ( t ) = log ⁡ 5 ( 7 t 2 + 4 ) g\left(t\right)=\log_5\left(7^{t^2+4}\right) g ( t ) = lo g 5 ( 7 t 2 + 4 )

2 t ln ⁡ 7 ln ⁡ 5 \frac{2t\ln7}{\ln5} l n 5 2 t l n 7

2 t ln ⁡ 7 5 2t\ln\frac75 2 t ln 5 7

2 t ln ⁡ 7 t ln ⁡ 5 \frac{2t\ln7}{t\ln5} t l n 5 2 t l n 7

7 t 2 + 4 t ln ⁡ 5 \frac{7^{t^{2+4}}}{t\ln5} t l n 5 7 t 2 + 4

4. Derivatives of Exponential & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

Business Calculus

4. Derivatives of Exponential & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

Struggling with Business Calculus?

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Multiple Choice

Find the derivative of the given function.
$g\left(t\right)=\log_5\left(7^{t^2+4}\right)$

$\frac{2t\ln7}{\ln5}$

$2t\ln\frac75$

$\frac{2t\ln7}{t\ln5}$

$\frac{7^{t^{2+4}}}{t\ln5}$

Verified step by step guidance

Step 1: Recognize that the function g(t) = log_5(7^(t^2 + 4)) involves a logarithm with a base other than e or 10. To differentiate this, we will use the change of base formula: log_a(x) = ln(x) / ln(a).

Step 2: Rewrite g(t) using the change of base formula: g(t) = ln(7^(t^2 + 4)) / ln(5). This allows us to work with natural logarithms, which are easier to differentiate.

Step 3: Simplify ln(7^(t^2 + 4)) using the logarithmic property ln(a^b) = b * ln(a). This gives g(t) = [(t^2 + 4) * ln(7)] / ln(5).

Step 4: Differentiate g(t) with respect to t. Use the quotient rule for derivatives: if h(t) = f(t) / g(t), then h'(t) = [f'(t)g(t) - f(t)g'(t)] / [g(t)]^2. Here, f(t) = (t^2 + 4) * ln(7) and g(t) = ln(5).

Step 5: Since ln(5) is a constant, its derivative is 0. Focus on differentiating f(t): f'(t) = d/dt[(t^2 + 4) * ln(7)]. Use the product rule: if u(t) = t^2 + 4 and v(t) = ln(7), then f'(t) = u'(t)v(t) + u(t)v'(t). Substitute the results into the quotient rule to find g'(t).

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