Applying the chain rule once more to the inner function e^(5x-3), the derivative of e^v is e^v · (dv)/(dx). Here v = 5x - 3, and its derivative with respect to x is 5.

For the second term, we rewrite (4)/(x) as 4x^(-1) and apply the power rule: (d)/(dx)(x^(-1)) = -1 · x^(-2) = -(1)/(x²). Multiplying by 4 yields -(4)/(x²).

2. Consider the function $f(x) = 3\cos(e^{5x-3}) + \frac{4}{x}$ Find $f'(x)$ $\left[\frac{d}{dx}(\cos x) = -\sin x\right]$

Answer: $f'(x) = -15e^{5x-3}\sin(e^{5x-3}) - \frac{4}{x^2}$

Step-by-step solution

Step 1: Apply linearity of differentiation

By the sum rule and constant multiple rule of differentiation, the derivative of a sum is the sum of the derivatives. We separate the expression into the trigonometric term and the rational term.

Step 2: Differentiate the first term using the chain rule

We use the chain rule for the composite function $3\cos(u)$ , where $u = e^{5x-3}$ . The derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$ , which we then multiply by the derivative of the inner function $u$ .

Step 3: Differentiate the inner exponential function

Applying the chain rule once more to the inner function $e^{5x-3}$ , the derivative of $e^v$ is $e^v \cdot \frac{dv}{dx}$ . Here $v = 5x - 3$ , and its derivative with respect to $x$ is $5$ .

Step 4: Differentiate the second term and combine results

For the second term, we rewrite $\frac{4}{x}$ as $4x^{-1}$ and apply the power rule: $\frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2} = -\frac{1}{x^2}$ . Multiplying by $4$ yields $-\frac{4}{x^2}$ .

Step 5: Combine and simplify to find $f'(x)$

Multiplying $3 \cdot (-\sin(e^{5x-3})) \cdot 5e^{5x-3}$ gives $-15e^{5x-3}\sin(e^{5x-3})$ . Combining this with the derivative of the second term gives the complete expression for $f'(x)$ .

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2. Consider the function f(x)=3cos⁡(e5x−3)+4xf(x) = 3\cos(e^{5x-3}) + \frac{4}{x}f(x)=3cos(e5x−3)+x4​ Find f′(x)f'(x)f′(x) [ddx(cos⁡x)=−sin⁡x]\left[\frac{d}{dx}(\cos x) = -\sin x\right][dxd​(cosx)=−sinx]