Applications of Derivatives
Product and Chain Rules
Grade 12
Question:
<p>If <math>y = \log x \cdot e^{(\tan x + x^2)}</math>, then <math>\frac{dy}{dx}</math> is equal to</p>
<p>(a) <math>e^{(\tan x + x^2)} \left[\frac{1}{x} + (\sec^2 x + 2x) \log x\right]</math></p>
<p>(b) <math>e^{(\tan x + x^2)} \left[\frac{1}{x} + (\sec^2 x - 2x) \log x\right]</math></p>
<p>(c) <math>e^{(\tan x + x^2)} \left[\frac{1}{x} + (\sec^2 x + 2x) \log x\right]</math></p>
<p>(d) <math>e^{(\tan x + x^2)} \left[\frac{1}{x} + (\sec^2 x - 2x) \log x\right]</math></p>
Step-by-Step Solution
Key Concept: We need to differentiate a product of two functions using the product rule, where one factor is logarithmic and the other is exponential with a composite exponent.
<p><strong>Step 1: Identify the structure</strong></p><p>We have y = (log x) · e^(tan x + x²), which is a product of two functions.</p><p>Let u = log x and v = e^(tan x + x²)</p><p><strong>Step 2: Apply the product rule</strong></p><p>dy/dx = u'v + uv'</p><p><strong>Step 3: Find u'</strong></p><p>u = log x</p><p>u' = 1/x</p><p><strong>Step 4: Find v'</strong></p><p>v = e^(tan x + x²)</p><p>Using the chain rule:</p><p>v' = e^(tan x + x²) · d/dx(tan x + x²)</p><p>v' = e^(tan x + x²) · (sec²x + 2x)</p><p><strong>Step 5: Substitute into product rule formula</strong></p><p>dy/dx = (1/x) · e^(tan x + x²) + (log x) · e^(tan x + x²) · (sec²x + 2x)</p><p><strong>Step 6: Factor out e^(tan x + x²)</strong></p><p>dy/dx = e^(tan x + x²) · [1/x + (log x)(sec²x + 2x)]</p><p>dy/dx = e^(tan x + x²) · [1/x + (sec²x + 2x)log x]</p><p><strong>∴ Answer:</strong> A</p>
Correct Answer: A