Q. Given the function y=2+x32x3, find dxdy in simplified form.Answer: dxdy=
Identify Function: Identify the function that needs to be differentiated.The function given is y=2+x32x3. We need to find the derivative of this function with respect to x, which is denoted as dxdy.
Apply Quotient Rule: Apply the quotient rule for differentiation. The quotient rule states that if we have a function y=vu, where both u and v are functions of x, then the derivative of y with respect to x is given by dxdy=v2vdxdu−udxdv. Here, u=2x3 and v=2+x3.
Differentiate u and v: Differentiate u and v with respect to x. The derivative of u=2x3 with respect to x is dxdu=6x2. The derivative of v=2+x3 with respect to x is v0.
Substitute Derivatives: Substitute the derivatives into the quotient rule formula.Using the derivatives from Step 3, we substitute into the quotient rule formula:dxdy=(2+x3)2(2+x3)(6x2)−(2x3)(3x2).
Simplify Expression: Simplify the expression.(dxdy)=4+4x3+x612x2+6x5−6x5.This simplifies to (dxdy)=4+4x3+x612x2.
Factor Out Common Terms: Further simplify the expression by factoring out common terms if possible.We can factor out a 4 from the numerator and denominator:dxdy=4×(1+x3+(41)x6)4×3x2.This simplifies to dxdy=1+x3+(41)x63x2.
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