12th Standard Mathematics — Differentials and Partial Derivatives: Online Practice Test

The area of a circle is \( A = \pi r^{2} \). Differentiating, \( dA = 2\pi r\,dr \), so the fractional error is \( \dfrac{dA}{A} = \dfrac{2\pi r\,dr}{\pi r^{2}} = \dfrac{2\,dr}{r} \). The percentage error in the area is therefore exactly twice the percentage error in the radius.

Here \( \dfrac{dr}{r} = \dfrac{0.02}{10} = 0.002 \), giving \( 2 \times 0.002 \times 100 = 0.4\% \). The squared dependence of area on radius is what doubles the error \(-\) a useful shortcut for area and volume problems.

Write \( y = x^{1/5} \). Taking the differential, \( dy = \tfrac{1}{5}\,x^{(1/5)-1}\,dx \), and dividing by \( y = x^{1/5} \) gives \( \dfrac{dy}{y} = \dfrac{1}{5}\cdot\dfrac{dx}{x} \).

So the relative \(-\) and hence percentage \(-\) error in the fifth root is one-fifth of that in the original number. In general the \(n\)th root scales the percentage error by \( \tfrac{1}{n} \); with \( n = 5 \) the factor is \( \tfrac{1}{5} \).

Differentiate partially with respect to \(x\), holding \(y\) constant. The exponent \( x^{2}+y^{2} \) has \(x\)-derivative \( 2x \), so by the chain rule \( \dfrac{\partial u}{\partial x} = e^{x^{2}+y^{2}}\cdot 2x = 2x\,e^{x^{2}+y^{2}} \).

Since \( u = e^{x^{2}+y^{2}} \) itself, this is simply \( 2xu \).

Differentiating \( v = \log(e^{x}+e^{y}) \) with respect to \(x\) (with \(y\) fixed) gives \( \dfrac{\partial v}{\partial x} = \dfrac{e^{x}}{e^{x}+e^{y}} \), since only the \( e^{x} \) term varies. By symmetry \( \dfrac{\partial v}{\partial y} = \dfrac{e^{y}}{e^{x}+e^{y}} \).

Adding them, the numerators combine to the full denominator: \( \dfrac{e^{x}+e^{y}}{e^{x}+e^{y}} = 1 \).

When differentiating \( x^{y} \) with respect to \(x\), the exponent \(y\) is held constant, so the expression behaves like an ordinary power function \( x^{n} \) with \( n = y \). The power rule then gives \( \dfrac{\partial w}{\partial x} = y\,x^{\,y-1} \).

Take care not to confuse this with differentiating with respect to \(y\): there \( x^{y} \) acts as an exponential and yields \( x^{y}\log x \).

First differentiate with respect to \(y\): \( \dfrac{\partial f}{\partial y} = x\,e^{xy} \). Now differentiate this with respect to \(x\) using the product rule on \( x\cdot e^{xy} \):

\[ \frac{\partial}{\partial x}\!\left(x\,e^{xy}\right) = e^{xy} + x\cdot y\,e^{xy} = (1+xy)\,e^{xy}. \]

The first term comes from differentiating the factor \(x\), the second from differentiating the exponential.

For a cube of edge \(x\), the volume is \( V = x^{3} \), so \( dV = 3x^{2}\,dx \). With \( x = 4 \) cm and \( dx = 0.1 \) cm,

\[ dV = 3(4)^{2}(0.1) = 3 \times 16 \times 0.1 = 4.8 \text{ cu.cm}. \]

The differential gives the first-order estimate of how far the computed volume can be off, given the measurement error in the edge.

The change in \(S\) is approximated by its differential. From \( S = 6x^{2} \), \( dS = 12x\,dx \); evaluating the derivative at \( x = x_{0} \) gives \( dS = 12x_{0}\,dx \).

Note the answer is a product (rate \( \times \) change), not a sum \(-\) options containing \( +\,dx \) wrongly mix the differential with the new edge length.

For \( V = x^{3} \), \( dV = 3x^{2}\,dx \). A \(1\%\) increase in the side means \( dx = 0.01x \). Substituting,

\[ dV = 3x^{2}(0.01x) = 0.03x^{3} \text{ m}^{3}. \]

The key step is converting the percentage increase into an actual increment \( dx = 0.01x \) before applying the differential.

Use the chain rule for a function of two variables that each depend on \(t\):

\[ \frac{dg}{dt} = \frac{\partial g}{\partial x}\frac{dx}{dt} + \frac{\partial g}{\partial y}\frac{dy}{dt}. \]

Here \( \dfrac{\partial g}{\partial x} = 6x \) and \( \dfrac{\partial g}{\partial y} = -5 + 4y \), while \( \dfrac{dx}{dt} = e^{t} \) and \( \dfrac{dy}{dt} = -\sin t \). Substituting \( x = e^{t},\ y = \cos t \):

\[ \frac{dg}{dt} = 6e^{t}\cdot e^{t} + (-5 + 4\cos t)(-\sin t) = 6e^{2t} + 5\sin t - 4\cos t\,\sin t. \]

Differentiate \( f(x) = \dfrac{x}{x+1} \) by the quotient rule:

\[ f'(x) = \frac{(x+1)\cdot 1 - x\cdot 1}{(x+1)^{2}} = \frac{1}{(x+1)^{2}}. \]

The differential is then \( df = f'(x)\,dx = \dfrac{1}{(x+1)^{2}}\,dx \). The numerator collapses to a constant because the \(x\)-terms cancel, leaving the squared denominator.

Differentiate partially with respect to \(x\), treating \(y\) and the constant \( -2019 \) as fixed: \( \dfrac{\partial u}{\partial x} = 2x + 3y \).

Evaluating at \( (x,y) = (4,-5) \): \( 2(4) + 3(-5) = 8 - 15 = -7 \). The standalone \(y\) term and the constant both vanish under \( \partial/\partial x \).

The linear approximation about \( x_{0} \) is \( L(x) = g(x_{0}) + g'(x_{0})(x - x_{0}) \). With \( x_{0} = \tfrac{\pi}{2} \): \( g\!\left(\tfrac{\pi}{2}\right) = \cos\tfrac{\pi}{2} = 0 \), and since \( g'(x) = -\sin x \), \( g'\!\left(\tfrac{\pi}{2}\right) = -1 \).

Therefore \( L(x) = 0 + (-1)\!\left(x - \tfrac{\pi}{2}\right) = -x + \tfrac{\pi}{2} \), which is the tangent line to \( \cos x \) at \( \tfrac{\pi}{2} \).

Compute each partial derivative, treating the other two variables as constants:

\[ w_{x} = 2x(y-z) - y^{2} + z^{2},\quad w_{y} = x^{2} + 2y(z-x) - z^{2},\quad w_{z} = -x^{2} + y^{2} + 2z(x-y). \]

Adding them, the quadratic terms \( -y^{2}+z^{2}+x^{2}-z^{2}-x^{2}+y^{2} \) cancel, and the cross terms \( 2x(y-z) + 2y(z-x) + 2z(x-y) = 2(xy - xz + yz - xy + zx - zy) = 0 \). Hence the sum is \(0\).

Do not confuse this plain sum with the weighted Euler sum \( x\,w_{x} + y\,w_{y} + z\,w_{z} \): \(w\) is homogeneous of degree 3, so that weighted sum equals \( 3w \), not zero.

The terms containing \(x\) are \(xy\) and \(zx\), so \( f_{x} = y + z \). The terms containing \(z\) are \(yz\) and \(zx\), so \( f_{z} = y + x \).

Subtracting, \( f_{x} - f_{z} = (y + z) - (y + x) = z - x \). The shared \(y\) cancels, leaving \( z - x \).